Leetcode 598 Range Addition II

Leetcode 598 Range Addition II

class Solution {public:    //memory limit exceeded    int maxCount(int m, int n, vector<vector<int>>& ops) {    vector<int> colMatrix(n,0);        vector<vector<int>> zeroMatrix(m,colMatrix);    int maxCount = 0;    //操作完之后统计与(0.0)元素相同的数    int row = 0;    int col = 0;    for(auto oneRow : ops)    {        row = oneRow[0];        col = oneRow[1];        for(int i = 0;i < row;i ++)        {            for(int j = 0;j < col;j ++)                zeroMatrix[i][j] ++;        }    }    for(int i = 0;i < m;i ++)    {        for(int j = 0;j < n;j ++)        {            if(zeroMatrix[i][j] == zeroMatrix[0][0])                maxCount ++;        }    }    return maxCount;    }    //找到ops中每个pair(a,b)的最小值即可，然后maxcount就是a*b    int maxCount2(int m, int n, vector<vector<int>>& ops) {    if(ops.empty())        return m*n;    int row = ops[0][0];    int col = ops[0][1];    for(auto oneRow : ops)    {        row = min(row,oneRow[0]);        col = min(col,oneRow[1]);    }    return row * col;    }};