leetcode 235-290 easy

235. Lowest Common Ancestor of a Binary Search Tree

公共的祖先必定大于左点小于右点，否则不断递归到合适。

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if ((root -> val > p -> val) && (root -> val > q -> val)) {
            return lowestCommonAncestor(root -> left, p, q);
        }
        if ((root -> val < p -> val) && (root -> val < q -> val)) {
            return lowestCommonAncestor(root -> right, p, q);
        }
        return root;
    }
};


/////////////////////////////////

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        TreeNode* cur = root;
        while (true) {
            if ((cur -> val > p -> val) && (cur -> val > q -> val)) {
                cur = cur -> left;
            } else if ((cur -> val < p -> val) && (cur -> val < q -> val)) {
                cur = cur -> right;
            } else {
                return cur;
            }
        }
    }
};

257. Binary Tree Paths

void binaryTreePaths(vector<string>& result, TreeNode* root, string t) {
    if(!root->left && !root->right) {
        result.push_back(t);
        return;
    }

    if(root->left) binaryTreePaths(result, root->left, t + "->" + to_string(root->left->val));
    if(root->right) binaryTreePaths(result, root->right, t + "->" + to_string(root->right->val));
}

vector<string> binaryTreePaths(TreeNode* root) {
    vector<string> result;
    if(!root) return result;
    
    binaryTreePaths(result, root, to_string(root->val));
    return result;
}

258. Add Digits

Iteration method

  class Solution(object):
  def addDigits(self, num):
    """
    :type num: int
    :rtype: int
    """
    while(num >= 10):
        temp = 0
        while(num > 0):
            temp += num % 10
            num /= 10
        num = temp
    return num
Digital Root

this method depends on the truth:

N=(a[0] * 1 + a[1] * 10 + ...a[n] * 10 ^n),and a[0]...a[n] are all between [0,9]

we set M = a[0] + a[1] + ..a[n]

and another truth is that:

1 % 9 = 1

10 % 9 = 1

100 % 9 = 1

so N % 9 = a[0] + a[1] + ..a[n]

means N % 9 = M

so N = M (% 9)

as 9 % 9 = 0,so we can make (n - 1) % 9 + 1 to help us solve the problem when n is 9.as N is 9, ( 9 - 1) % 9 + 1 = 9

///就是一个数的数根等于该数各位数的和的mod 9
///  (num-1)%9+1  等于 num%9，这为了解决9的树根时9而不是0的问题
class Solution(object):
def addDigits(self, num):
    """
    :type num: int
    :rtype: int
    """
    if num == 0 : return 0
    else:return (num - 1) % 9 + 1

263. Ugly Number

bool isUgly(int num) {
    if(num == 0) return false;
    
    while(num%2 == 0) num/=2;
    while(num%3 == 0) num/=3;
    while(num%5 == 0) num/=5;
    
    return num == 1;
}

268. Missing Number

1.XOR
相同则为0。num[]的数字和下标一样
public int missingNumber(int[] nums) { //xor
    int res = nums.length;
    for(int i=0; i<nums.length; i++){
        res ^= i;
        res ^= nums[i];
    }
    return res;
}
2.SUM
public int missingNumber(int[] nums) { //sum
    int len = nums.length;
    int sum = (0+len)*(len+1)/2;
    for(int i=0; i<len; i++)
        sum-=nums[i];
    return sum;
}
3.Binary Search
public int missingNumber(int[] nums) { //binary search
    Arrays.sort(nums);
    int left = 0, right = nums.length, mid= (left + right)/2;
    while(left<right){
        mid = (left + right)/2;
        if(nums[mid]>mid) right = mid;
        else left = mid+1;
    }
    return left;
}
Summary:
If the array is in order, I prefer Binary Search method. Otherwise, the XOR method is better.

283. Move Zeroes

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int j = 0;
        // move all the nonzero elements advance
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] != 0) {
                nums[j++] = nums[i];
            }
        }
        for (;j < nums.size(); j++) {
            nums[j] = 0;
        }
    }
};

290. Word Pattern

Input: pattern = "abba", str = "dog cat cat dog"
Output: true

bool wordPattern(string pattern, string str) {
    map<char, int> p2i;
    map<string, int> w2i;
    istringstream in(str);
    int i = 0, n = pattern.size();
    for (string word; in >> word; ++i) {
        if (i == n || p2i[pattern[i]] != w2i[word])
            return false;
        p2i[pattern[i]] = w2i[word] = i + 1;
    }
    return i == n;
}