高盛昂赛 算法题先写corner case

【方法】

字写大点，先注释框架

链表：指针走就行了，最多是两个同时一起走。

两个链表求交点

//corner case

        if (headA == null || headB == null) {
            return null;
        }
        //keep the same length
        int A_len = getLength(headA);
        int B_len = getLength(headB);
        
        while (A_len > B_len) {
            headA = headA.next;
            A_len--;
        }
        while (A_len < B_len) {
            headB = headB.next;
            B_len--;
        }
        //find the same node
        while (headA != headB) {
            headA = headA.next;
            headB = headB.next;
        }

LCA:用两个点和要找的方向来DC

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
        if (root == null || A == root || B == root) {//
            return root;
        }
        //divide
        TreeNode left = lowestCommonAncestor(root.left, A, B);
        TreeNode right = lowestCommonAncestor(root.right, A, B);
        
        //conquer
        if (left != null && right != null) {
            return root;//
        }
        else if (left != null) {
            return left;
        }
        else if (right != null) {
            return right;
        }
        else {
            return null;
        }
    }

 

 

 

diameter of tree树的周长

 

public int depth(TreeNode root) {
        //corner case
        if (root == null) {
            return 0;
        }
        //define left, right
        int left = depth(root.left);
        int right = depth(root.right);
        //renew max, don't add 1 since it's a sum of two lines
        max = Math.max(max, left + right);
        //return val for root, notice
        return Math.max(left, right) + 1;
    }

 对称的树：先判断相等

class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        //corner case
        if (s == null) {
            return false;
        }
        if (isSame(s,t)) {
            return true;
        }
        return isSubtree(s.left, t) || isSubtree(s.right, t);
    }
    
    public boolean isSame(TreeNode s, TreeNode t) {
        //both null
        if (s == null && t == null) {
            return true;
        }
        //one is null
        if (s == null || t == null) {
            return false;
        }
        //false
        if (s.val != t.val) {
            return false;
        }
        //default
        return isSame(s.left, t.left) && isSame(s.right, t.right);
    }
}

 同时要判断平衡和深度，自定义新的数据类型：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */


public class Solution {
    /*
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
     class ResultType {
       boolean isBalanced;
       int maxDepth;
       ResultType(boolean isBalanced,int maxDepth) {
           this.isBalanced = isBalanced;
           this.maxDepth = maxDepth;
       }
     };
     
    public boolean isBalanced(TreeNode root) {
        return helper(root).isBalanced;
    }
    
    private ResultType helper (TreeNode root) {
        if (root == null) {
            return new ResultType(true, 0);
        }
        
        ResultType left = helper(root.left);
        ResultType right = helper(root.right);
        
        if (!left.isBalanced || !right.isBalanced) {
            return new ResultType(false, - 1);
        }
        
        else if (Math.abs(left.maxDepth - right.maxDepth) > 1) {
            return new ResultType(false, - 1);
        }
        else {
            return new ResultType(true, Math.max(left.maxDepth,right.maxDepth) + 1);
        }
    }
}

Binary Tree Paths 用iteration

void findBT(TreeNode root, String path, List<String> ans) {
        if (root.left == null && root.right == null) 
            ans.add(path + root.val);//add here since it's an sign of end
        if (root.left != null) findBT(root.left, path + root.val + "->"参数是path,每次新加一个节点, ans);
        if (root.right != null) findBT(root.right, path + root.val + "->", ans);
    }

merge binary tree

class Node 
{ 
    int data; 
    Node left, right; 
      
    public Node(int data, Node left, Node right) { 
        this.data = data; 
        this.left = left; 
        this.right = right; 
    } 
      
     /* Helper method that allocates a new node with the 
     given data and NULL left and right pointers. */
     static Node newNode(int data) 
     { 
         return new Node(data, null, null); 
     } 
       
     /* Given a binary tree, print its nodes in inorder*/
     static void inorder(Node node) 
     { 
         if (node == null) 
             return; 
        
         /* first recur on left child */
         inorder(node.left); 
        
         /* then print the data of node */
         System.out.printf("%d ", node.data); 
        
         /* now recur on right child */
         inorder(node.right); 
     } 
       
     /* Method to merge given two binary trees*/
     static Node MergeTrees(Node t1, Node t2) 
     { 
         if (t1 == null) 
             return t2; 
         if (t2 == null) 
             return t1; 
         t1.data += t2.data; 
         t1.left = MergeTrees(t1.left, t2.left); 
         t1.right = MergeTrees(t1.right, t2.right); 
         return t1; 
     } 

 

217. Contains Duplicate

class Solution {
    public boolean containsDuplicate(int[] nums) {
        //cc
        if (nums.length == 0 || nums == null) {
            return false;
        }
        //ini = sort
        Arrays.sort(nums);记得先排序
        //for 
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] == nums[i - 1]) {
                直接return true;
            }
        }
        return false;
    }
}

 

 

Excel Sheet Column Number

class Solution {
    public int titleToNumber(String s) {
        int res = 0;
        for (int i = 0; i < s.length(); i++) {
　　　　　　　　26进制
            res = res * 26 + (s.charAt(i) - 'A' + 1);
        }
        return res;
    }
}

2的倍数，三种方法：

Last Edit: October 22, 2018 8:24 PM

motorix
motorix
 260
This question is not an difficult one, and there are many ways to solve it.

Method 1: Iterative

check if n can be divided by 2. If yes, divide n by 2 and check it repeatedly.

if (n == 0) return false;
while (n%2 == 0) n/=2;
return n == 1;
Time complexity = O(log n)

Method 2: Recursive

return n > 0 && (n == 1 || (n%2 == 0 && isPowerOfTwo(n/2)));
Time complexity = O(log n)

Method 3: Bit operation

If n is the power of two:

n = 2 ^ 0 = 1 = 0b0000...00000001, and (n - 1) = 0 = 0b0000...0000.
n = 2 ^ 1 = 2 = 0b0000...00000010, and (n - 1) = 1 = 0b0000...0001.
n = 2 ^ 2 = 4 = 0b0000...00000100, and (n - 1) = 3 = 0b0000...0011.
n = 2 ^ 3 = 8 = 0b0000...00001000, and (n - 1) = 7 = 0b0000...0111.
we have n & (n-1) == 0b0000...0000 == 0

Otherwise, n & (n-1) != 0.

For example, n =14 = 0b0000...1110, and (n - 1) = 13 = 0b0000...1101.

return n > 0 && ((n & (n-1)) == 0);
Time complexity = O(1)

 

二进制中1的个数 往右移动获得数字

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int ones = 0;
        while (n != 0) {
            ones += (n & 1);
            n = n >>> 1;
        }
        return ones;
    }
}

 

3．变换符号
变换符号就是正数变成负数，负数变成正数。

 

如对于-11和11，可以通过下面的变换方法将-11变成11

 

      1111 0101(二进制) –取反-> 0000 1010(二进制) –加1-> 0000 1011(二进制)

 

同样可以这样的将11变成-11

 

      0000 1011(二进制) –取反-> 0000 0100(二进制) –加1-> 1111 0101(二进制)

 

因此变换符号只需要取反后加1即可

 

找两个字符串的区别：用异或就对了

 

class Solution {
    public char findTheDifference(String s, String t) {
    char c = 0;
    for (int i = 0; i < s.length(); ++i) {
        c ^= s.charAt(i);
    }
    for (int i = 0; i < t.length(); ++i) {
        c ^= t.charAt(i);
    }
    return c;
}
}

 

 

 

连续自然数数组中缺失的数字 ：异或2次

//268. Missing Number
//for loop, judge if nums[i + 1] != nums[i] + 1
class Solution {
public int missingNumber(int[] nums) { //xor
    int res = nums.length;
    System.out.println(" res = " + res);
    
    for(int i=0; i<nums.length; i++){
        System.out.println(" i = " + i);
        System.out.println(" nums[i] = " + nums[i]);
        
        res ^= i;
        System.out.println(" res ^= i = " + res);
        res ^= nums[i];
        System.out.println(" res ^= nums[i] = " + res);
        System.out.println(" ");
    }
    return res;
}
}
//n 1

 

 

 

191. Number of 1 Bits 往右移动

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int ones = 0;
        while (n != 0) {
　　　　　　　计算每一位数与1的结果
            ones += (n & 1);
            n = n >>> 1;
        }
        return ones;
    }
}

104. Maximum Depth of Binary Tree 最基本的DC了，DC就记住这个就行

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: An integer.
     */
    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        int left = maxDepth(root.left);
        int right = maxDepth(root.right);
        
        int max = Math.max(left,right);
        return max + 1;
    }
}

 

108. Convert Sorted Array to Binary Search Tree

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        //TreeNode node = new TreeNode(0);
        //corner case
        if (nums.length == 0) {
            return null;
        }
        TreeNode node = helper(nums, 0, nums.length - 1);// -1 should be noticed ahead
        return node;
    }
    必须要有三个参数，不同的函数才能被称作helper
    public TreeNode helper(int[] nums, int low, int high) {      
    //corner case : low > high
        if (low > high) {
            return null;
        }
        int mid = low + (high - low) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = helper(nums, low, mid - 1);
        root.right = helper(nums, mid + 1, high);
        要用到函数里给的参数
        return root;
    }
}

 

118. Pascal's Trianglclass Solution    

public List<List<Integer>> generate(int numRows)

//ini
        List<List<Integer>> triangle = new ArrayList<List<Integer>>();
        
        //cc
        if (numRows <= 0) return triangle; 
            
        //for
        for (int i = 0; i < numRows; i++) {
            List<Integer> row = new ArrayList<Integer>();
            
            for (int j = 0; j < i + 1; j++) {
　　　　　　　　　　//首先写corner case
                if (j == 0 || j == i) {
                    row.add(1);
                }else {
                    row.add(triangle.get(i - 1).get(j - 1) + triangle.get(i - 1).get(j));
                }     
            }
            //add again
　　　　　　　　
用add方法
                triangle.add(new ArrayList(row));
        }
            
        //return
        return triangle;
    }
}

 

plusOne 数组加一，从最后开始进位

public class Solution {
    /**
     * @param digits: a number represented as an array of digits
     * @return: the result
     */
    public int[] plusOne(int[] digits) {
        //not carry
        int n = digits.length;
        //0 to n-1 whenever
        for (int i = n - 1; i >= 0; i--) {
            if (digits[i] < 9) {
                digits[i]++;
                return digits;直接在这里return即可
            }else {
                digits[i] = 0;
            }
        }
        //carry, need new array
        int[] answer = new int[n + 1];
        answer[0] = 1;
        return answer;
    }
}

 

deleteNode

public class Solution {
    /*
     * @param node: the node in the list should be deletedt
     * @return: nothing
     */
    public void deleteNode(ListNode node) {
        // write your code here
        if (node == null || node.next == null) {
            return ;
        }
注意删除的是next而不是当前的node 
        ListNode next = node.next;
        node.val = next.val;
        node.next = next.next;
    }
}

 

190. Reverse Bits

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        //ini res
        int res = 0;
        
        //for loop: 32 32位数，做移动32次
        for (int i = 0; i < 32; i++) {
            res <<= 1; 左移留出最后一位为0来相加
            if ((n & 1) == 1) res += 1;
            n >>= 1; 右移留出最后一位来操作
        }
        
        return res;
    }
}

 

101. Symmetric Tree

 

class Solution {
    public boolean isSymmetric(TreeNode root) {
        //corner case
        if (root == null) {
            return true;
        }
        return isSymmetricHelper(root.left, root.right);
    }
    
    public boolean isSymmetricHelper(TreeNode left, TreeNode right) {
        //all null
        if (left == null && right == null) {
            return true;
        }
        //one null
        if (left == null || right == null) {
            return false;
        }
        //not same 
        if (left.val != right.val) {
            return false;
        }
        //same
        return isSymmetricHelper(left.left, right.right) && isSymmetricHelper(left.right, right.left);
        //don't forget the function name
    }
}

 

 

 

234. Palindrome Linked List 两边开始

 

public class Solution {
    /**
     * @param head: A ListNode.
     * @return: A boolean.
     */
    public boolean isPalindrome(ListNode head) {
        // corner case
        if (head == null) {
            return true;//true
        }
        //p1 go with p2
        ListNode middle = findMiddle(head);
        middle.next = reverse(middle.next);//
        
        ListNode p1 = head, p2 = middle.next;
        while (p2 != null && p1.val == p2.val) {//order
            p1 = p1.next;
            p2 = p2.next;
        }
        return p2 == null;
    }
    
    private ListNode findMiddle(ListNode head) {
        // corner case
        if (head == null) {
            return null;
        }无处不在的corner case
        ListNode fast = head.next, slow = head; 先走一步
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    
    private ListNode reverse(ListNode head) {
        // corner case
        if (head == null) {
            return null;
        }
        ListNode prev = null;
        while (head != null) {
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }
        return prev;
    }
}

 

 

 

136. Single Number 出现一次 用异或

 

class Solution {
    public int singleNumber(int[] nums) {
        //ini n
        int n = 0;
        
        //for loop ^=
        for (int i = 0; i < nums.length; i++) {
            n ^= nums[i]; 不同2次会返回原来的值
        }
        
        //return
        if (n != 0) return n;
        
        return 0;
    }
}

 

 

 

169. Majority Element（n/2 times）

 

class Solution {
    public int majorityElement(int[] nums) {
        //ini
        int major = nums[0];
        int count = 1;
        //1 - n
        for (int i = 1; i < nums.length; i++) {
            //nums[i] != major, count--,if count == 0
            if (nums[i] != major) {
                if (count == 0) {
                    major = nums[i];
                    count++;
                }
                count--;抵消当前的
            }else {
                //nums[i] == major, count++
                count++;
            }    
        }
        return major;
    }
}

160. Intersection of Two Linked Lists 想想图，对着写就行了

 

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

 

 

public class Solution {
    /*
     * @param headA: the first list
     * @param headB: the second list
     * @return: a ListNode
     */
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        //corner case
        if (headA == null || headB == null) {
            return null;
        }
        //keep the same length
        int A_len = getLength(headA);
        int B_len = getLength(headB);
        
        while (A_len > B_len) {
            headA = headA.next;
            A_len--;
        }
        while (A_len < B_len) {
            headB = headB.next;
            B_len--;
        }
        //find the same node
        while (headA != headB) {
            headA = headA.next;
            headB = headB.next;
        }
        
        return headA;
    }
    
    //getLength
    public int getLength(ListNode node) {
        int length = 0;
        while(node != null) {
            length++;
            node = node.next;
        }
        return length;
    }
}

26. Remove Duplicates from Sorted Array

 

class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int size = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != nums[size]) { 总是去和size代表的最后一位相比较
                nums[++size] = nums[i];
            }
        }
        return size + 1;剩一个0位
    }
}

 

387. First Unique Character in a String 

public class Solution {
    /**
     * @param s: a string
     * @return: it's index
     */
    public int firstUniqChar(String s) {
        //corner case
        if (s == null) {
            return 0;
        }
        //put into cnt[]
        char[] c = s.toCharArray();字母是一个单独的数组
        int[] cnt = new int[256];表示所有字符
        for (int i = 0; i < s.length(); i++) {
            cnt[c[i]]++;
        }
        //return
        for (int i = 0; i < s.length(); i++) {
            if (cnt[c[i]] == 1) {
                return i;
                //break;
            }
        }
        return -1;
    }
}

 

14. Longest Common Prefix

 

Input: ["flower","flow","flight"]
Output: "fl"

 

public class Solution {
    /**
     * @param strs: A list of strings
     * @return: The longest common prefix
     */
    public String longestCommonPrefix(String[] strs) {
        //corner case
        if (strs == null) {
            return "";
        }
        if (strs.length == 0) {
            return "";
        }
        //define pre
        String pre = strs[0];
        int n = strs.length;
        //shorten pre
        for (int i = 1; i < n; i++) {
            while (strs[i].indexOf(pre) != 0) {
                pre = pre.substring(0, pre.length() - 1);
            }
        }
        //return
        return pre;
    }
}

 

283. Move Zeroes

class Solution {
    public void moveZeroes(int[] nums) {
        //cc
        if (nums == null || nums.length == 0) return;
        //ini
        int insertPos = 0;
 
        for (int num : nums) {
            //no 0s
            if (num != 0) {
                nums[insertPos++] = num;
            }
        }
         //0s 分开就行
            while (insertPos < nums.length) {
                nums[insertPos++] = 0;
            }
        //return
    }
}

 

//88. Merge Sorted Array
//add the bigger number from backward 从后往前

public class Solution {
    /*
     * @param A: sorted integer array A which has m elements, but size of A is m+n
     * @param m: An integer
     * @param B: sorted integer array B which has n elements
     * @param n: An integer
     * @return: nothing
     */
    public void merge(int[] A, int m, int[] B, int n) {
        // write your code here
        int index = m + n - 1;
        int i = m - 1;
        int j = n - 1;
        //add the bigger number from backward
        while (i >= 0 && j >= 0) {最重要的是这一步
            if (A[i] > B [j]) {
                A[index--] = A[i--];
            }
            else {
                A[index--] = B[j--];
            }
        }
        //if there's number remaning in just A or B, append to the result
        while (i >= 0) {
                A[index--] = A[i--];
            }
        while (j >= 0) {
                A[index--] = B[j--];
            }
    }
}

 

 

Merge Two Sorted Lists 从前往后

 

public class Solution {
    /*
     * @param l1: ListNode l1 is the head of the linked list
     * @param l2: ListNode l2 is the head of the linked list
     * @return: ListNode head of linked list
     */
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        // write your code here
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;得用dummy node
        while(l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                tail.next = l1;
                l1 = l1.next;
            }
            else {
                tail.next = l2;
                l2 = l2.next;
            }
            tail = tail.next;一样的，注意要有主变量
        }
        if (l1 != null) {
            tail.next = l1;
        }
        if (l2 != null) {
            tail.next = l2;
        }
        return dummy.next;
    }
}

 

 

 

121. Best Time to Buy and Sell Stock

//121. Best Time to Buy and Sell Stock
//use greedy to solve
class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length == 0 || prices == null) {
            return 0;
        }
        
        int min = Integer.MAX_VALUE;
        int profit = 0;
        //update the min price and the max profit each time
        for (int i = 0; i < prices.length; i++) {
　　　　　　两个三元运算符
            min = min < prices[i] ? min : prices[i];
            profit没有i = (prices[i] - min) > profit ? prices[i] - min : profit;
        }
        
        return profit;
    }
}
//n 1

 

Reverse Integer: 我不熟悉的对每一位数进行的操作，步骤都是：取数、迭代、改数

public int reverse(int x) {
        //ini
        int res = 0, newRes = 0;
        
        while (x != 0) {
            int tail = x % 10;先余后除，中间是newres 先背吧
            newRes = res * 10 + tail;
            if ((newRes - tail) / 10 != res) {
                return 0;
            }
            res = newRes;
            x = x / 10;
        }
        
        //return 
        return newRes;
    }

 

Maximum Subarray

class Solution {
    public int maxSubArray(int[] nums) {
        if (nums.length == 0 || nums == null) {
            return -1;
        }
        
        int sum = 0;
        int minSum = 0;
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            max = Math.max(max, sum - minSum);先减之前的
            minSum = Math.min(minSum, sum);
        }
        
        return max;
    }
}

 

 

 

Two Sum

 

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        
        for (int i = 0; i < nums.length; i++) {
            if (map.containsKey(target - nums[i])) {
                result[0] = map.get(target - nums[i]);模板index
                result[1] = i;
                
                return result;
            }
            map.put(nums[i],i);
        }
        return result;
    }
}

Linked List Cycle

public class Solution {
    /*
     * @param head: The first node of linked list.
     * @return: True if it has a cycle, or false
     */
    public boolean hasCycle(ListNode head) {
        // write your code here
        if (head == null) {
            return false;
        }
        ListNode fast = head.next;
        ListNode slow = head;
        while(fast != slow) {
            if (fast == null || fast.next == null) {
                return false;
            }
            fast = fast.next.next;
            slow = slow.next;
        }
        
        return true;
    }
}

 

Add Two Numbers 从前往后加，carry在最后面

public class Solution {
    /*
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2 
     */
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        //intialize
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;//
        //sum
        int carry = 0;
        for (ListNode i = l1, j = l2; i != null || j != null;) {
            int sum = carry;
            sum += (i != null)? i.val: 0;注意一下是否为空再加
            sum += (j != null)? j.val: 0;
            // %
            tail.next = new ListNode(sum % 10);
            tail = tail.next;
            // /
            carry = sum / 10;
            i = (i == null)? i : i.next;//i, j should go down
            j = (j == null)? j : j.next;//judge itself
        }
        //carry 用于最后结尾收尸
        if (carry != 0) {
            tail.next = new ListNode(carry);
        }//turn carry into a node
        
        return dummy.next;//forget
    }
}

 

万年不会的翻转链表

 

class Solution {
    public ListNode reverseList(ListNode head) {
        
        ListNode prev = null;
        ListNode curt = head;
        
        while(curt != null) {
            ListNode temp = curt.next;
            curt.next = prev;
            prev = curt;
            curt = temp;
        }
        return prev;
    }
}

 

 

 

 

1. 找两个array不同的element: 就用双重for循环，没有啥高级的办法，就这样！

public static void main(String[] args) {
        // TODO Auto-generated method stub

        int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6 };
        int[] arr2 = new int[] { 5, 6, 7, 8 };

        boolean contains = false;
        List<Integer> list = new ArrayList<Integer>();
        for (int i = 0; i < arr1.length; i++) {
            for (int j = 0; j < arr2.length; j++) {
                if (arr1[i] == arr2[j]) {
                    contains = true;
                    break;
                }
            }

            if(!contains){
                list.add(arr1[i]);
            }
            else{
                contains = false;
            }
        }
        System.out.println(list);

    }

2. 怎么实现hashmap，big O是多少

 key- index - bucket[index] array

 

private int getBucketIndex(K key) 
    { 
        int hashCode = key.hashCode(); 
        int index = hashCode % numBuckets; 
        return index; 
    } 

 

好像不一样，不用看了

public class Book implements Comparable<Book>{
String title, author; int year;
Book(String title, String author, int year) {
//the usual stuff
}
@Override
public int compareTo(Book b) {
return (this.title.compareTo(b.title));
}
@Override
public int hashCode() {  
return  Objects.hash(title, author, year);
}
@Override
public boolean equals(Object o) { 
if (o == null) return false;      
if (this == o) return true;    
if (getClass() != o.getClass()) return false; 
Book b = (Book) o; 
return title.equals(b.title) 
        && author.equals(b.author) 
        && (year == b.year); 
}}

 

3. friend circle

class Solution {
    public int findCircleNum(int[][] M) {
        //corner case
        if (M == null || M.length == 0) return 0;
        
        //initialization: count = n, each id = id
        int m = M.length;
        int count = m;
        int[] roots = new int[m];
        for (int i = 0; i < m; i++) roots[i] = i; 
        
        //for loop and union find
        for (int i = 0; i < m; i++) {
            for (int j = i + 1; j < m; j++) {
                //if there is an edge, do union find
                if (M[i][j] == 1) {
//注意下是怎么来的
                    int root0 = find (roots, i);
                    int root1 = find (roots, j);
                    
                    if (root0 != root1) {
                        roots[root1] = root0;
                        count--;
                    }
                }
            }
        }
        
        //return count
        return count;
    }
    
    public int find (int[] roots, int id) {
        while (id != roots[id]) {
            id = roots[roots[id]];
        }
        return id;
    }
}

 

4. 找array倒数第k个element；找array连续element前面的element

 只让我写了一个sql和一个merge sort...我觉得那不算算法题

mergesort: 新建两个数组，recursively merge之后再拼起来

//recursive and append
public static void mergeSort(int[] a, int n) {
    if (n < 2) {
        return;
    }
    int mid = n / 2;
    int[] l = new int[mid];
//不一定是双倍，要用n-mid
    int[] r = new int[n - mid];
 
    for (int i = 0; i < mid; i++) {
        l[i] = a[i];
    }
    for (int i = mid; i < n; i++) {
        r[i - mid] = a[i];
//for loop里就不用++了
    }
//sort - merge
    mergeSort(l, mid);
    mergeSort(r, n - mid);
 
    merge(a, l, r, mid, n - mid);
}

public static void merge(
  int[] a, int[] l, int[] r, int left, int right) {
  //list 5 variables here
    int i = 0, j = 0, k = 0;
    while (i < left && j < right) {
        if (l[i] < r[j]) {
            a[k++] = l[i++];
        }
        else {
            a[k++] = r[j++];
        }
    }
//should be outside
    while (i < left) {
        a[k++] = l[i++];
    }
    while (j < right) {
        a[k++] = r[j++];
    }
}

quicksort:移动index, index还是原来的i < j就得交换后继续
package com.java2novice.sorting;
 
public class MyQuickSort {
     
    private int array[];
    private int length;
 
    public void sort(int[] inputArr) {
         
        if (inputArr == null || inputArr.length == 0) {
            return;
        }
        this.array = inputArr;
        length = inputArr.length;
        quickSort(0, length - 1);
    }
 
    private void quickSort(int lowerIndex, int higherIndex) {
         
        int i = lowerIndex;
        int j = higherIndex;
        // calculate pivot number, I am taking pivot as middle index number
        int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
pivot应该是数组中的一个元素
        // Divide into two arrays
//写完i j的关系之后要写整个数组的recursive
        while (i <= j) {
            /**
             * In each iteration, we will identify a number from left side which 
             * is greater then the pivot value, and also we will identify a number 
             * from right side which is less then the pivot value. Once the search 
             * is done, then we exchange both numbers.
             */
            while (array[i] < pivot) {
                i++;
            }
            while (array[j] > pivot) {
                j--;
            }
//array[i] is bigger than pivot, so exchange and continue
            if (i <= j) {
                exchangeNumbers(i, j);
                //move index to next position on both sides
                i++;
                j--;
            }
        }
        // call quickSort() method recursively
        if (lowerIndex < j)
            quickSort(lowerIndex, j);
        if (i < higherIndex)
            quickSort(i, higherIndex);
    }
 
    private void exchangeNumbers(int i, int j) {
        int temp = array[i];
        array[i] = array[j];
        array[j] = temp;
    }
     
    public static void main(String a[]){
         
        MyQuickSort sorter = new MyQuickSort();
        int[] input = {24,2,45,20,56,75,2,56,99,53,12};
        sorter.sort(input);
        for(int i:input){
            System.out.print(i);
            System.out.print(" ");
        }
    }
}

 

说给100首歌，设计一个shuffle/repeat播放音乐的逻辑.本文原创自1point3acres论坛
没有写代码，就说了说思路，也并不在乎你用啥语言，我用的python问能不能用buildin func 或者numpy都允许了hhh.

就用numpy random一个整数范围是0～n然后每次放了这个歌以后直接把这条数据移到最后然后n-1就行，最后n=1的时候播放最后一首歌，同时n再置最开始的那个数值就好。感觉这题是现场出的，说是考数据结构我一开始想用map，还迂回了很久= =。希望不要觉得我笨

 

【valid parenthesis】
class Solution {
public boolean isValid(String s) {
//corner case
if (s == null) {
return true;
}
Stack<Character> stack = new Stack<Character>();
//push into stack
for (char c : s.toCharArray()) {
//left cases
if (c == '(') {
stack.push(')');
}else if (c == '[') {

//每一种都要写
stack.push(']');
}else if (c == '{') {
stack.push('}');
}else if (stack.isEmpty() || stack.pop() != c) {
//right case
return false;
}
}
return stack.isEmpty();
}
}

【invert binary tree】
class TreeNode {
int val; TreeNode left, right; //parameter is another item
TreeNode(int item) { val = item; left = right = null; }}
class Solution {
public TreeNode invertTree(TreeNode root) {

//careful with corner case
if (root == null) {
return null;
}
TreeNode temp = root.left;
root.left = invertTree(root.right);
root.right = invertTree(temp);
return root;
}
}

【给两个String，找相同的字符】就是双重for循环
public static void main(String[] args) {

 

String str1 = "刘烨,孙坚,王二小,蜘蛛侠,钢铁侠,毛剑卿";
String str2 = "王二小,李占军,刘胡兰,毛剑卿";

 

String[] arr1 = str1.split(",") ;
String[] arr2 = str2.split(",") ;
StringBuffer sb = new StringBuffer();
for (int i = 0; i < arr2.length; i++){
for (int j = 0; j < arr1.length; j++){
if (arr1[j].equals(arr2[i])){
sb.append(arr1[j] + ",") ;
}
}
}
System.out.println("结果：" + sb.toString().substring(0, sb.toString().length() - 1));

}
【reverse vowels】典型双指针了
class Solution {
public String reverseVowels(String s) {
//cc
if (s == null) {
return s;
}
//ini
String vowels = "aeiouAEIOU";
char[] chars = s.toCharArray();
int start = 0, end = s.length() - 1;

while (start < end) {
//adjust start, end
while (start < end && !vowels.contains(chars[start] + "")) {
start++;
}
while (start < end && !vowels.contains(chars[end] + "")) {
end--;
}
//exchange
char temp = chars[start];
chars[start] = chars[end];
chars[end] = temp;
//push to move on
start++;
end--;
}

//return
return new String(chars);
}
}

【missing number】短路了吧？
class Solution {
public int missingNumber(int[] nums) {

//ini = sort
Arrays.sort(nums);

//cc
if (nums[0] != 0) {
return 0;
}

//for
for (int i = 0; i < nums.length - 1; i++) {
//notice
if (nums[i + 1] != nums[i] + 1) {
return nums[i] + 1;
}
}

//return
return nums.length;
}
}

【reverse string】
class Solution {
public String reverseString(String s) {
//corner case
if (s == null) {
return null;
}
int i = 0, j = s.length() - 1;
//convert to char[]
char[] chars = s.toCharArray();
while (i < j) {
char temp = chars[i];
chars[i] = chars[j];
chars[j] = temp;
//notice here
i++;
j--;
}
//convert again
return new String(chars);
//return
}
}

【方法2】
public class Solution {
public String reverseString(String s) {
return new StringBuilder(s).reverse().toString();
}
}

【Valid Palindrome】
public class Solution {
/**
* @param s: A string
* @return: Whether the string is a valid palindrome
*/
public boolean isPalindrome(String s) {
//cc
if (s.isEmpty()) {
return true;
}

//define
int head = 0, tail = s.length() - 1;

while (head < tail) {
char cHead = s.charAt(head), cTail = s.charAt(tail);
//judge:punction or not
if (!Character.isLetterOrDigit(cHead)) {
head++;
}else if (!Character.isLetterOrDigit(cTail)) {
tail--;
}else {
//tolowercase and judge
if (Character.toLowerCase(cHead) != Character.toLowerCase(cTail)) {
return false;
}
//continue
head++;
tail--;
}
}

//return
return true;
}

【二分法解根号2】
class Solution {
public int mySqrt(int x) {
//bs
long start = 1, end = x;
while (start + 1 < end) {
long mid = start + (end - start) / 2;
if (mid * mid <= x) {
start = mid;
}else {
end = mid;
}
}

//return end or start
if (end * end <= x) {
return (int)end;
}
return (int)start;
}
}

 

【斐波那契数列 recursion】
//Fibonacci Series using Recursion
class fibonacci
{
static int fib(int n)
{
if (n <= 1)
return n;
return fib(n-1) + fib(n-2);
}

public static void main (String args[])
{
int n = 9;
System.out.println(fib(n));
}
}
Time Complexity: T(n) = T(n-1) + T(n-2) which is exponential.

【Use Dynamic Programming】
// Fibonacci Series using Dynamic Programming
class fibonacci
{
static int fib(int n)
{
/* Declare an array to store Fibonacci numbers. */
int f[] = new int[n+2]; // 1 extra to handle case, n = 0
int i;

/* 0th and 1st number of the series are 0 and 1*/
f[0] = 0;
f[1] = 1;

for (i = 2; i <= n; i++)
{
/* Add the previous 2 numbers in the series
and store it */
f[i] = f[i-1] + f[i-2];
}

return f[n];
}

public static void main (String args[])
{
int n = 9;
System.out.println(fib(n));
}
}
//n n

【井字棋】
public class TicTacToe {
private int[] rows;
private int[] cols;
private int diagonal;
private int antiDiagonal;

/** Initialize your data structure here. */
public TicTacToe(int n) {
rows = new int[n];
cols = new int[n];
}

/** Player {player} makes a move at ({row}, {col}).
@param row The row of the board.
@param col The column of the board.
@param player The player, can be either 1 or 2.
@return The current winning condition, can be either:
0: No one wins.
1: Player 1 wins.
2: Player 2 wins. */
public int move(int row, int col, int player) {
int toAdd = player == 1 ? 1 : -1;

rows[row] += toAdd;
cols[col] += toAdd;
if (row == col)
{
diagonal += toAdd;
}

if (col == (cols.length - row - 1))
{
antiDiagonal += toAdd;
}

int size = rows.length;
if (Math.abs(rows[row]) == size ||
Math.abs(cols[col]) == size ||
Math.abs(diagonal) == size ||
Math.abs(antiDiagonal) == size)
{
return player;
}

return 0;
}
}

【happy number】
class Solution {
public boolean isHappy(int n) {
//cc
if (n == 0) {
return false;
}

//ini
int squareSum, remain;
Set set = new HashSet();

//while loop, contains

//每个n 在set中只加一次，不重复
while (set.add(n)) {
squareSum = 0;
//每次加一个n,都重新算squareSum
while (n > 0) {
remain = n % 10;
squareSum += remain * remain;
n = n / 10;
}

if (squareSum == 1) return true;
n = squareSum;最后才把数字给n
}

return false;
}
}

【number permutation】
public class Solution {
/*
* @param nums: A list of integers.
* @return: A list of permutations.
*/
public List<List<Integer>> permute(int[] nums) {
//corner case
List<List<Integer>> results = new ArrayList<>();//
List<Integer> permutations = new ArrayList<Integer>();
HashSet<Integer> set = new HashSet<Integer>();

if (nums == null) {
return results;
}
if (nums.length == 0) {
results.add(new ArrayList<Integer>());
return results;
}
//helper
helper(nums, permutations, set, results);
return results;
}

//helper ：NSTRS
public void helper(int[] nums, int start, List<Integer> temp, List<List<Integer>> results,HashSet<Integer> set用于检验重复) {
if (permutations.size() == nums.length) {//exit
results.add(new ArrayList<Integer>(permutations));
return ; 记得此处要返回，有return
}

for (int i = 0; i < nums.length; i++) {

//continue when there is duplicate 注意重复的情况，这也是一种corner case
if (set.contains(nums[i])) {
continue;
}

temp.add(nums[i]);
set.add(nums[i]);
helper(nums, i+1, temp, results, set);
set.remove(nums[i]);
temp.remove(temp.size() - 1);
}
}

位运算

public ArrayList<String> permute(char[] s, ArrayList<String> list, int index) {

        list.add(String.valueOf(s));
        if (index >= s.length)
            return list;

        for (int i = index; i < s.length; i++) {

            if (Character.isAlphabetic(s[i])) {

                s[i] ^= 32;
                permute(s, list, i + 1);
                s[i] ^= 32;
            }
        }
        return list;

    }

 

subset:

private void backtrack(int [] nums, int start,List<Integer> tempList, List<List<Integer>> list){
list.add(new ArrayList<>(tempList));
for(int i = start; i < nums.length; i++){
tempList.add(nums[i]);
backtrack(list, tempList, nums, i + 1);
tempList.remove(tempList.size() - 1);
}
}

 

【search in 2 2D array】
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {

if (matrix == null || matrix.length == 0) {
return false;
}
if (matrix[0] == null || matrix[0].length == 0) {
return false;
}

int m = matrix.length;
int n = matrix[0].length;
int x = m - 1;
int y = 0;
int count = 0;

while (x >= 0 && y < n) {
if (matrix[x][y] > target) {

大动
x--;
}
else if(matrix[x][y] < target) {

小动

y++;
}
else {
count++;
x--;
y++;
}
}

if (count > 0) {
return true;
}
return false;
}
}

【group anagrams】
class Solution {
public List<List<String>> groupAnagrams(String[] strs) {
//cc
if (strs == null || strs.length == 0) return new ArrayList<List<String>>();

//ini: map, char[]
Map<String, List<String>> map = new HashMap<>();

//for loop: add to char, to map
for (String str : strs) {
char[] chars = str.toCharArray();

//顺序统一以后都一样
Arrays.sort(chars);
String anagram = String.valueOf(chars);

//then add to map
if (!map.containsKey(anagram)) map.put(anagram, new ArrayList<String>());
map.get(anagram).add(str);
}

//return (map)
return new ArrayList<List<String>>(map.values());
}
}

【find second maximum】固定2个数
// JAVA Code for Find Second largest
// element in an array
class GFG {

/* Function to print the second largest
elements */
public static void print2largest(int arr[],
int arr_size)
{
int i, first, second;

/* There should be atleast two elements */
if (arr_size < 2)
{
System.out.print(" Invalid Input ");
return;
}

first = second = Integer.MIN_VALUE;
for (i = 0; i < arr_size ; i++)
{
/* If current element is smaller than
first then update both first and second */
if (arr[i] > first)
{
second = first;
first = arr[i];
}

/* If arr[i] is in between first and
second then update second */
else if (arr[i] > second && arr[i] != first)
second = arr[i];
}

if (second == Integer.MIN_VALUE)
System.out.print("There is no second largest"+
" element ");
else
System.out.print("The second largest element"+
" is "+ second);
}

【Find All Duplicates in an Array】

class Solution {
public int removeDuplicates(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int size = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != nums[size]) {

//compare with the last number
nums[++size] = nums[i];

//注意是先加再给
}
}
return size + 1;
}
}

【difference】
比较基础 递归 迭代：本身、堆栈、基础条件
基本 函数体中的语句调用函数本身。 允许重复执行指令集。
格式 在递归函数中，仅指定终止条件（基本情况）。 迭代包括初始化，条件，循环内语句的执行以及控制变量的更新（递增和递减）。
终止 条件语句包含在函数体中，以强制函数返回而不执行递归调用。 重复执行迭代语句，直到达到某个条件。
条件 如果函数没有收敛到某个称为（基本情况）的条件，则会导致无限递归。 如果迭代语句中的控制条件永远不会变为false，则会导致无限次迭代。
无限重复 无限递归可能会导致系统崩溃。 无限循环重复使用CPU周期。
应用的 递归始终应用于函数。 迭代应用于迭代语句或“循环”。
堆 每次调用函数时，堆栈用于存储一组新的局部变量和参数。 不使用堆栈。
高架 递归拥有重复函数调用的开销。 没有重复函数调用的开销。
速度 执行缓慢。 快速执行。
代码大小 递归减少了代码的大小。 迭代使代码更长。

【】