代码随想录32|738.单调递增的数字，968.监控二叉树，56. 合并区间

738.单调递增的数字

链接地址

class Solution {
public:
    int monotoneIncreasingDigits(int n) {
        string str = to_string(n);
        int flag = str.size();
        for (int i = str.size() - 1; i > 0; i--) {
            if (str[i] < str[i - 1]) {
                str[i - 1] -= 1;
                flag = i;
            }
        }

        for (int j = flag; j < str.size(); j++) {
            str[j] = '9';
        }

        return stoi(str);

    }
};
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968.监控二叉树

链接地址

/**
 * 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 {
    
private:
    int Tranversal(TreeNode* root, int& result) {
        if (root == nullptr) return 2;
        int left = Tranversal(root->left, result);
        int right = Tranversal(root->right, result);
        if (left == 2 && right == 2) {
            return 0;
        }
        //情况1：
        if (left == 0 || right == 0) {
            result++;
            return 1;
        }
        if (left == 1 || right == 1) {
            return 2;
        } 
        return -1;
        
    }
public:
    int minCameraCover(TreeNode* root) {
        int result = 0;
        if (Tranversal(root, result) == 0) {
            result++;
        }
        return result;
    }
};
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56. 合并区间

链接地址

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        vector<vector<int>> result;
        if (intervals.size() == 0) return result;
        sort(intervals.begin(), intervals.end(), [](const vector<int>& a, const vector<int>& b){return a[0] < b[0];});
        result.push_back(intervals[0]);
        for (int i = 0; i < intervals.size(); i++) {
            if (result.back()[1] >= intervals[i][0]) {
                result.back()[1] = max(intervals[i][1], result.back()[1]); 
            } else {
                result.push_back(intervals[i]);
            }
        }
        return result;
    }
};
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