代码随想录算法训练营第三十七天|738.单调递增的数字、968.监控二叉树

738.单调递增的数字

题目链接：. - 力扣（LeetCode）
文档讲解：代码随想录
视频讲解：贪心算法，思路不难想，但代码不好写！LeetCode:738.单调自增的数字_哔哩哔哩_bilibili

C++代码:

class Solution {
public:
    int monotoneIncreasingDigits(int n) {
        string strNum = to_string(n);
        int flag = strNum.size();
        for (int i = strNum.size() - 1; i > 0; i--) {
            if (strNum[i - 1] > strNum[i] ) {
                flag = i;
                strNum[i - 1]--;
            }
        }
        for (int i = flag; i < strNum.size(); i++) {
            strNum[i] = '9';
        }
        return stoi(strNum);
    }
};

968.监控二叉树

题目链接：. - 力扣（LeetCode）
文档讲解：代码随想录
视频讲解：贪心算法，二叉树与贪心的结合，有点难...... LeetCode:968.监督二叉树_哔哩哔哩_bilibili

C++代码:

/**
 * 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) {
        if (cur == NULL) return 2;
        int left = traversal(cur->left);
        int right = traversal(cur->right);
        if (left == 2 && right == 2) return 0;
        if (left == 0 || right == 0) {
            result++;
            return 1;
        }
        if (left == 1 || right == 1) return 2;
        return -1;
    }
    int minCameraCover(TreeNode* root) {
        result = 0;
        if (traversal(root) == 0) {
            result++;
        }
        return result;
    }
};

总结

摘自 代码随想录知识星球 (opens new window)成员：海螺人