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;
}
};