代码随想录算法训练营第三十七 | ● 738.单调递增的数字 ● 968.监控二叉树

738.单调递增的数字

讲解链接：https://programmercarl.com/0738.%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E7%9A%84%E6%95%B0%E5%AD%97.html

class Solution {
public:int monotoneIncreasingDigits(int n) {//整数转字符串，变为字符串访问比诸位取出数字要好。string strNum = to_string(n);int flag = strNum.size();for(int i=strNum.size()-1;i>=0;i--) {//找到从哪个位置赋值为9if(strNum[i-1]>strNum[i]) {//i-1处减一，从i往后都赋为9flag = i;strNum[i-1] --;}}for(int i=flag;i<strNum.size();i++) {strNum[i]='9';}//将字符串转为十进制return stoic(strNum);}
};

968.监控二叉树

讲解链接：https://programmercarl.com/0968.%E7%9B%91%E6%8E%A7%E4%BA%8C%E5%8F%89%E6%A0%91.html

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:int result;int traversal(TreeNode* cur) {//将空结点看作是有覆盖（标记为2）if(cur==NULL)return 2;int left = traversal(cur->left);int right = traversal(cur->right);//左右节点都有覆盖if(left == 2 && right ==2)return 0;//左右至少有一个节点没有被覆盖else if(left==0 || right==0) {result++;return 1;}elsereturn 2;}int minCameraCover(TreeNode* root) {result = 0;if(traversal(root) ==0)result++;return result;}};