LeetCode_22_Apr_4th_Week

真真是天有不测风云，中间亲人出了点事情就没有精力继续刷题了。万幸是这个时间节点还可以多照顾她，现在情况也稳定下来了，环境也差不多配置好了，再捡回来好了。祝大家都健健康康，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

主要是随机种子的设置和使用：

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;
};

评论这里有个蓄水池原理，挺有意思的，效率也比上述方法更高

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

该题就是面积为，每列最大值的和+每行最大值的和+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; //复用行列的遍历，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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