LeetCode_22_Apr_4th_Week

It's really unpredictable,If something happened to the relatives in the middle, there is no energy to continue to brush the questions.Fortunately, this time node can take more care of her,The situation has now stabilized,The environment is almost configured,Pick it up again.I wish you all good health,生活如意！

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April 25th : 398. 随机数索引
April 26th : 883. 三维形体投影面积

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April 25th : 398. 随机数索引

Mainly the setting and use of random seeds：

srand((unsigned)time(NULL));
rand()%num //等概率获取0~num-1之间的任一数字

解题代码如下：

class Solution {
    
public:
    Solution(vector<int>& nums) {
    
        srand((unsigned)time(NULL));
        for(int i = 0; i < nums.size(); ++i) keys[nums[i]].push_back(i);
    }
    
    int pick(int target) {
    
        return keys[target][rand()%keys[target].size()];
    }
private:
    unordered_map<int, vector<int>> keys;
};

Comment There is a reservoir principle here,挺有意思的,The efficiency is also higher than the above method

class Solution {
    
public:
    vector<int> res;
    Solution(vector<int>& nums) {
    
       res = nums; 
    }
    
    int pick(int target) {
    
        int c = 0, index = 0;
        for(int i = 0;i < res.size();i++)
            if(res[i] == target){
    
                c++;
                if(rand() % c == 0) index = i;
            }
        return index;
    }
};

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April 26th :April 26th : 883. 三维形体投影面积

The problem is that the area is ,The sum of the maximum values ​​for each column+The sum of the maximum values ​​for each row+非0行列数目,解题代码如下：

class Solution {
    
public:
    int projectionArea(vector<vector<int>>& grid) {
    
        int overLookView = 0, leftLookView = 0, rightLookView = 0;
        int rowMax = 0, colMax = 0; //Traversal of multiplexed rows and columns,只需交换i,j的位置即可
        for(int i = 0; i < grid.size(); ++i) {
    
            rowMax = 0, colMax = 0;
            for(int j = 0; j < grid[i].size(); ++j) {
    
                overLookView += grid[i][j] ? 1 : 0;
                rowMax = max(grid[i][j], rowMax);
                colMax = max(grid[j][i], colMax);
            }
            leftLookView += rowMax;
            rightLookView += colMax;
        }
        return overLookView + leftLookView + rightLookView;
    }
};

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