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Problem: 807. 保持城市天际线

解题过程

耗时100%，拿到每一行每一列的最大值，能够添加的数值，就是该数字所在行i最大值所在列j最大值的最小值 - grid[i][j]，sum += min(rowMX[i], colMX[j]) - grid[i][j];

Code

class Solution { public: int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) { int n = grid.size(); vector<int> rowMX(n, INT_MIN), colMX(n, INT_MIN); for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { rowMX[i] = max(rowMX[i], grid[i][j]); colMX[i] = max(colMX[i], grid[j][i]); } } int sum = 0; for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { sum += min(rowMX[i], colMX[j]) - grid[i][j]; } } return sum; } };