面试题3 二维数组中的查找 Leetcode--74 Search a 2D Matrix
1 /*Java 2 Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: 3 4 Integers in each row are sorted from left to right. 5 The first integer of each row is greater than the last integer of the previous row. 6 7 For example, 8 9 Consider the following matrix: 10 11 [ 12 [1, 3, 5, 7], 13 [10, 11, 16, 20], 14 [23, 30, 34, 50] 15 ] 16 17 Given target = 3, return true. 18 本题最简单的方法循环遍历行与列 但是时间复杂度较高。 19 */ 20 class Solution { 21 public boolean searchMatrix(int[][] matrix, int target) { 22 boolean find = false; 23 int rows,cols; 24 if (matrix == null || matrix.length==0) { 25 rows = 0; 26 cols = 0; 27 } else { 28 rows = matrix.length; 29 cols = matrix[0].length; 30 } 31 32 if (matrix != null && rows > 0 && cols > 0) { 33 int row = 0; 34 int col = cols-1; 35 while(row<rows && col>=0){ 36 if(matrix[row][col]==target){ 37 find=true; 38 break; 39 } 40 41 else if(matrix[row][col]>target){ 42 col--; 43 }else{ 44 row++; 45 } 46 } 47 } 48 49 return find; 50 } 51 public static void main(String[] args) { 52 int[][] matrix1 = { { 1, 3, 5, 7 }, { 10, 11, 16, 20 }, { 23, 30, 34, 50 } }; 53 54 int[][] matrix = {}; 55 int target = 55; 56 SearchA2DMatrix cc = new SearchA2DMatrix(); 57 boolean find = cc.searchMatrix(matrix, target); 58 System.out.println(find); 59 } 60 }
面试题5 从尾到头打印链表 -------Leetcode 206题 Reverse Linked List
package easy;
import java.util.Stack;
import easy.ListNode;
public class L206ReverseLinkedList {
/*
* Reverse a singly linked list. 先读入，后输出。典型的先进后出，可用栈来实现。
*/
/**
* Definition for singly-linked list.
*
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public ListNode reverseList(ListNode head) {
if (head == null)
return head;
Stack<Integer> myStack = new Stack<Integer>();
ListNode pNode = head;
while (pNode != null) {
myStack.push(pNode.val);
pNode = pNode.next;
}
int size = myStack.size();
ListNode newHead = new ListNode(myStack.peek());
myStack.pop();
if (size == 1)
return newHead;
ListNode node = new ListNode(myStack.peek());
newHead.next = node;
myStack.pop();
while (!myStack.empty()) {
node.next = new ListNode(myStack.peek());
myStack.pop();
node = node.next;
}
while (newHead != null) {
System.out.println(newHead.val);
newHead = newHead.next;
}
return newHead;
}
public static void main(String[] args) {
ListNode head1 = new ListNode(1);
ListNode head2 = new ListNode(2);
ListNode head3 = new ListNode(3);
ListNode head4 = new ListNode(4);
head1.next = head2;
head2.next = head3;
head3.next = head4;
head4.next = null;
L206ReverseLinkedList cc = new L206ReverseLinkedList();
cc.reverseList(head1);
}
}
面试题6:重建二叉树，由二叉树的前序遍历和中序遍历重建二叉树——Leetcode 105.Construct Binary Tree from Preorder and Inorder Traversal
package medium;
//面试题6
public class L105ConstructBinaryTreefromPreorderandInorderTraversal {
/*
* 根据前序遍历与中序遍历重建一个二叉树
*/
public TreeNode buildTree(int[] preorder, int[] inorder) {
if (preorder == null || inorder == null) {
return null;
}
int length = preorder.length;
if(length==0){
TreeNode root =null;
return root;
}
//System.out.println(length);
int startPre = 0;
int endPre = length - 1;
int startIn = 0;
int endIn = length - 1;
return buildTreeCore(preorder, inorder, startPre, endPre, startIn, endIn);
}
public TreeNode buildTreeCore(int[] preorder, int[] inorder, int startPre, int endPre, int startIn, int endIn) {
// 前序遍历的第一个结点为根节点。
int rootValue=preorder[startPre];
TreeNode root = new TreeNode(rootValue);
root.left = root.right = null;
System.out.println("前序遍历第一个节点"+preorder[startPre]);
if (preorder[startPre] == preorder[endPre] )
if(inorder[startIn] == inorder[endIn]) {
return root;
} else {
System.out.println("非法输入");
}
//在中序遍历中找到根节点
int rootInorderIndex=startIn;
while(startIn<=endIn && inorder[rootInorderIndex]!=rootValue){
rootInorderIndex++;
}
int leftLen=rootInorderIndex-startIn;
int leftPreEnd=startPre+leftLen;
if(leftLen>0){
root.left=buildTreeCore(preorder,inorder,startPre+1,leftPreEnd,startIn,rootInorderIndex-1);
}
if(leftLen<endPre-startPre){
root.right=buildTreeCore(preorder,inorder,leftPreEnd+1,endPre,rootInorderIndex+1,endIn);
}
return root;
}
public static void main(String[] args) {
L105ConstructBinaryTreefromPreorderandInorderTraversal cc= new L105ConstructBinaryTreefromPreorderandInorderTraversal();
int[] preorder1={1,2,4,7,3,5,6,8};
int[] inorder1={4,7,2,1,5,3,8,6};
int[] preorder={1,2};
int[] inorder={2,1};
cc.buildTree(preorder, inorder);
}
}
面试题6-2:重建二叉树，由二叉树的中序遍历和后序遍历重建二叉树——Leetcode 106.Construct Binary Tree from Inorder and Postorder Traversal
package medium;
public class L106ConstructBinaryTreefromInorderandPostorderTraversal {
/*
* 根据前序遍历与中序遍历重建一个二叉树
*/
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (postorder == null || inorder == null) {
return null;
}
int length = inorder.length;
if (length == 0) {
TreeNode root = null;
return root;
}
// System.out.println(length);
int startPost = 0;
int endPost = length - 1;
int startIn = 0;
int endIn = length - 1;
return buildTreeCore(inorder, postorder, startIn, endIn, startPost, endPost);
}
public TreeNode buildTreeCore(int[] inorder, int[] postorder, int startIn, int endIn, int startPost, int endPost) {
// 后序遍历的最后一个结点为根节点。
int rootValue = postorder[endPost];
TreeNode root = new TreeNode(rootValue);
root.left = root.right = null;
if (postorder[startPost] == postorder[endPost]) {
if (inorder[startIn] == inorder[endIn]) {
return root;
} else {
System.out.println("非法输入");
}
}
// 在中序遍历中找到根节点
int rootInorderIndex = startIn;
while (startIn <= endIn && inorder[rootInorderIndex] != rootValue) {
rootInorderIndex++;
}
int leftLen = rootInorderIndex - startIn;
int leftPostEnd = startPost + leftLen-1;
if (leftLen > 0) {
root.left = buildTreeCore(inorder, postorder, startIn, rootInorderIndex - 1, startPost, leftPostEnd);
}
if (leftLen < endPost - startPost) {
root.right = buildTreeCore(inorder, postorder, rootInorderIndex + 1, endIn, leftPostEnd + 1, endPost - 1);
}
return root;
}
public static void main(String[] args) {
L106ConstructBinaryTreefromInorderandPostorderTraversal cc = new L106ConstructBinaryTreefromInorderandPostorderTraversal();
int[] inorder = { 4, 7, 2, 1, 5, 3, 8, 6 };
int[] postorder={ 7, 4, 2, 5, 8, 6, 3, 1 };
cc.buildTree( inorder,postorder);
}
}
面试题7 用两个栈，实现队列-----Leetcode 232题 Implement Queue using Stacks
package easy;
import java.util.Stack;
public class L232ImplementQueueUsingStacks {
Stack<Integer> queue = new Stack();
public void push(int x) {
Stack<Integer> stack = new Stack();
while (!queue.empty()) {
stack.push(queue.pop());
}
queue.push(x);
while (!stack.empty()) {
queue.push(stack.pop());
}
}
public int pop() {
return queue.pop();
}
public int peek() {
return queue.peek();
}
public boolean empty() {
return queue.empty();
}
public static void main(String[] args) {
MyQueue obj = new MyQueue();
obj.push(2);
obj.push(1);
obj.push(3);
obj.push(4);
int param_2 = obj.pop();
int param_3 = obj.peek();
boolean param_4 = obj.empty();
System.out.println("pop:"+param_2);
System.out.println("peek:"+param_3);
System.out.println("empty:"+param_4);
}
}
面试题7-2 用两个队列，实现栈-----L225题 Implement Stacks using Queue
package easy;
import java.util.LinkedList;
import java.util.Queue;
public class L225_ImplementStackUsingQueues {
/*
* 用两个栈实现队列 先入先出
*
* Implement the following operations of a stack using queues.
*/
Queue<Integer> queue1 = new LinkedList<Integer>();
Queue<Integer> queue2 = new LinkedList<Integer>();
// push(x) -- Push element x onto stack.
public void push(int x) {
if (queue1.isEmpty() && queue2.isEmpty()) {
queue1.add(x);
} else if (!queue1.isEmpty()) {
queue1.add(x);
} else {
queue2.add(x);
}
}
// pop() -- Removes the element on top of the stack.
public int pop() {
int size1 = queue1.size();
int size2 = queue2.size();
if (size1 > 0) {
while (size1 > 1) {
queue2.add(queue1.poll());
size1--;
}
return queue1.poll();
} else if (size2 > 0) {
while (size2 > 1) {
queue1.add(queue2.poll());
size2--;
}
return queue2.poll();
} else {
System.out.println("栈为空");
}
return 0;
}
// top() -- Get the top element.
public int top() {
int size1 = queue1.size();
int size2 = queue2.size();
int x=0;
if (size1 > 0) {
while (size1 > 1) {
queue2.add(queue1.poll());
size1--;
}
x=queue1.peek();
queue2.add(queue1.poll());
} else if (size2 > 0) {
while (size2 > 1) {
queue1.add(queue2.poll());
size2--;
}
x=queue2.peek();
queue1.add(queue2.poll());
} else {
System.out.println("栈为空");
}
return x;
}
// empty() -- Return whether the stack is empty.
public boolean empty() {
int size = queue1.size() + queue2.size();
return size > 0 ? false : true;
}
public static void main(String[] args) {
//["MyStack","push","push","push","top",
//"pop","top","pop","top","empty","pop","empty"]
L225_ImplementStackUsingQueues obj = new L225_ImplementStackUsingQueues();
obj.push(1);
System.out.println("top1:" +obj.top());
obj.push(2);
System.out.println("top2:" +obj.top());
obj.push(3);
System.out.println("top3:" +obj.top());
System.out.println("pop:" +obj.pop() );
System.out.println("top:" +obj.top());
System.out.println("pop:" +obj.pop() );
System.out.println("top:" +obj.top());
System.out.println("empty:" + obj.empty());
System.out.println("pop:" +obj.pop() );
System.out.println("empty:" + obj.empty());
}
}
面试题8:旋转数组的最小数字—leetcode 153题 Find Minimum in Rotated Sorted Array
class Solution {
public int findMin(int[] nums) {
int min=nums[0];
int len=nums.length;
int low=0;
int high=len-1;
int mid=(low+high)/2;
if(nums[low]<nums[high]){
return nums[low];
}
while(low<=high){
if(high-low==1 || high==low){
System.out.println("min:"+min);
return nums[high]<=nums[high] ? nums[high]:nums[low];
}
if(nums[mid]>nums[high]){
min=nums[high];
low=mid;
}else{
min=nums[mid];
high=mid;
}
mid=(low+high)/2;
}
return min;
}
}
面试题9：菲波那切数列
package jianzhiOffer;
public class I9Fibonacci {
public int calFi1(int n) {
// 递归
if (n == 0) {
return 0;
}
if (n == 1) {
return 1;
}
return calFi1(n - 1) + calFi1(n - 2);
}
public int calFi2(int n) {
// 循环
if (n == 0) {
return 0;
}
if (n == 1) {
return 1;
}
int one = 0;
int two = 1;
int sum = 0;
for (int i = 2; i <= n; i++) {
sum = one + two;
one = two;
two = sum;
}
return sum;
}
public static void main(String[] args) {
I9Fibonacci cc = new I9Fibonacci();
int sum1 = cc.calFi1(3);
int sum2 = cc.calFi2(3);
System.out.println(sum1 + ":" + sum2);
}
}
面试题10:二进制中1的个数——剑指OfferP78 ~~~L191题 Number of 1 Bits
package easy;
public class L191NumberOf1Bits {
// you need to treat n as an unsigned value
//将一个整数减去一个1，这个整数的二进制变化是：最右面的1变为0，这一位之后的所有位变为1.
//如1100 减去一为1011.
//减1之后的整数与原来的整数做与运算，会把原整数最右面一位变为0.
//1011 & 1100=1000
public int hammingWeight(int n) {
int number=0;
while(n!=0){
number++;
n=(n-1)&n;
}
return number;
}
//解法2 较慢的解法，如果n&1结果为1这表示n的最后一位为1，count++;之后将1左移，判断前面位。
public int solu2(int n){
int number=0;
int flag=1;
while(flag!=0){
if((n & flag) !=0){
number++;
}
flag=flag<<1;
}
return number;
}
public static void main(String[] args) {
L191NumberOf1Bits cc= new L191NumberOf1Bits();
int number=cc.hammingWeight(2);
System.out.println(number);
}
}
面试题11:数值的整数次方——Leetcde 50.Pow(x, n) 详解
面试题14:调整数组顺序使奇数位于偶数前面——Leetcode 328. Odd Even Linked List 详解
面试题15:链表中倒数第k个结点~~~~Leetcode 19. Remove Nth Node From End of List.详解
面试题16:反转链表~~~~~Leetcode 92. Reverse Linked List II 详解
Leetcode 206.Reverse Linked List 详解
面试题17:合并两个排序的链表~~~~~Leetcode 21. Merge Two Sorted Lists 详解
面试题20:顺时针打印矩阵~~~~Leetcode 54.Spiral Matrix 详解
59题 Spiral MatrixII 详解
面试题25:二叉树中和为某一值的路径~~~~Leetcode 112 Path Sum 详解
Leetcode 113 PathSumII 详解
面试题29:数组中出现次数超过一半的数字~~~~~Leetcode 169.Majority Element 详解
面试题33:把数组排成最小的数~~~~~Leetcode 179.Largest Number 详解
面试题37：两个链表中的第一个公共节点~~~~~Leetcode 160. Intersection of Two Linked Lists 详解
面试题39：二叉树的深度~~~~~Leetcode 110. Balanced Binary Tree 详解
面试题50：把字符串变成整数（atoi函数）~~~~Leetcode 8.String to Integer (atoi) 详解
面试题51：树中两个结点的最低公共祖先~~~~~~Leetcode 235. Lowest Common Ancestor of a Binary Search Tree详解
Leetcode 236.Lowest Common Ancestor of a Binary Tree 详解