807. Max Increase to Keep City Skyline

问题描述：

In a 2 dimensional array grid, each value grid[i][j] represents the height of a building located there. We are allowed to increase the height of any number of buildings, by any amount (the amounts can be different for different buildings). Height 0 is considered to be a building as well. 

At the end, the "skyline" when viewed from all four directions of the grid, i.e. top, bottom, left, and right, must be the same as the skyline of the original grid. A city's skyline is the outer contour of the rectangles formed by all the buildings when viewed from a distance. See the following example.

What is the maximum total sum that the height of the buildings can be increased?

Example:
Input: grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
Output: 35
Explanation: 
The grid is:
[ [3, 0, 8, 4], 
  [2, 4, 5, 7],
  [9, 2, 6, 3],
  [0, 3, 1, 0] ]

The skyline viewed from top or bottom is: [9, 4, 8, 7]
The skyline viewed from left or right is: [8, 7, 9, 3]

The grid after increasing the height of buildings without affecting skylines is:

gridNew = [ [8, 4, 8, 7],
            [7, 4, 7, 7],
            [9, 4, 8, 7],
            [3, 3, 3, 3] ]

Notes:

• 1 < grid.length = grid[0].length <= 50.
• All heights grid[i][j] are in the range [0, 100].
• All buildings in grid[i][j] occupy the entire grid cell: that is, they are a 1 x 1 x grid[i][j] rectangular prism.

解题思路：

首先遍历整个矩阵存储来得到每一行，每一列的最大值。

每一行的最大值存储在rowMax中，每一列的最大值存储在columnMax中。

注意边界情况:

　　当grid.size() == 0 || grid[0].size() == 0时，显然没有东西可以修改; 

　　当grid.size() == 1 || grid[0].size() == 1时，为了保证自个方向看没有改动，所以不能进行改动。

然后在遍历整个矩阵，注意在点grid[i][j]的最大值应该为min(rowMax[i], columnMax[j])

代码：

class Solution {
public:
    int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        if(m == 0 || n == 0 || m == 1 || n == 1) return 0;
        vector<int> rowMax(m, 0);
        vector<int> columnMax(n, 0);
        
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                rowMax[i] = max(rowMax[i], grid[i][j]);
                columnMax[j] = max(columnMax[j], grid[i][j]);
            }
        }
        
        int ret = 0;
        
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                int mx = min(rowMax[i], columnMax[j]);
                ret += (mx - grid[i][j]);
            }
        }
        return ret;
    }
};