2021/8/24

leetcode3个困难

看了场景题10题；

两场笔试

mysql 5讲

图解网络

 

 十道海量数据处理面试题与十个方法大总结_知识库_博客园 (cnblogs.com)

[//]: # (支持粘贴图片啦🎉🎉🎉)
[//]: # (保存的笔记可以在小程序中查看)

```

class Solution {
public:
    int dp[100100];
    int candy(vector<int>& ratings) {
        int L = 0,R = 1,ans = 0,pre = 0;
        pre = 1;ans += 1;
        for(int i = 1; i < ratings.size(); i ++){
            if(ratings[i] >= ratings[i - 1]){
                L = 0; 
                if(ratings[i] == ratings[i - 1]) {
                    pre = 1; 
                } else pre += 1;
                ans += pre;
                R = pre;
            } else {
                L ++;
                if(L == R){
                    L ++;
                }
                ans += L;
                pre = 1;
            }
        }
        return ans;
    }
};
```

　　

```
//笔记将同步在小程序中，便于复习
/**
 * 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 maxx = -1010;
    inline int dfs(TreeNode *root){
        if(root == NULL) return 0;
        int left = dfs(root -> left);
        int right = dfs(root -> right);
        maxx = max(maxx,root -> val);
        maxx = max(maxx,left + right + root -> val);
        return max(max(max(left,right) + root -> val,0),root -> val);
    }
    int maxPathSum(TreeNode* root) {
         dfs(root); return maxx;
    }
};
```