[LeetCode] Max Increase to Keep City Skyline

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.

为保持天际线不变，楼房最多可以增长的层数。

思路如下：

1. 先找出grid中横纵两条天际线。存放在数组中。题中示例如下

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

2. 按照天际线的标准来增加建筑的高度。以x[0]为标准，如果y[i]小于x[0]，则以y[i]为最高高度。如果y[i]大于x[0]，则以x[0]为最高高度。然后将对应高度与原高度求差即可。

class Solution {
public:
    int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
        if (grid.empty())
            return 0;
        int m = grid.size();
        int n = grid[0].size();
        vector<int> rows(m, 0);
        vector<int> cols(n, 0);
        for (int i = 0; i < m; ++i)
        {
            int maxX = 0;
            for (int j = 0; j < n; ++j)
            {
                maxX = max(maxX, grid[i][j]);
            }
            rows[i] = maxX;
        }
        for (int j = 0; j < m; ++j)
        {
            int maxY = 0;
            for (int i = 0; i < n; ++i)
            {
                maxY = max(maxY, grid[i][j]);
            }
            cols[j] = maxY;
        }
        int cnt = 0;
        for (int i = 0; i < m; ++i)
        {
            int tmp = rows[i];
            for (int j = 0; j < n; ++j)
            {
                if (cols[j] > tmp)
                {
                    cnt += (tmp - grid[i][j]);
                }
                else
                {
                    cnt += (cols[j] - grid[i][j]);
                }
            }
        }
        return cnt;
    }
};