807. 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.

题目大意： 给定二维数组。 grid[i][j] 表示 第i行，第j列的楼高度。原始的二维数组中，每一行的最大值为left or right的skyline。 每一列的最大值 为top to bottom 的skyline。 各个grid 的元素可以 增加高度，但是不能破坏 left2right 以及top2bottom的skyline。

算法：straight forward

class Solution {
    public int maxIncreaseKeepingSkyline(int[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        int res = 0;
        int rowNumber = grid.length;
        int columnNumber = grid[0].length;
        int[] top2Bottom = new int[columnNumber];
        int[] left2Right = new int[rowNumber];
        //找出从上到下，每一列中最大的值
        for (int i = 0; i < top2Bottom.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                top2Bottom[i] = Math.max(top2Bottom[i], grid[i][j]);
            }
        }
        //找出从左到右，每一行最大值
        for (int i = 0; i < left2Right.length; i++) {
            for (int j = 0; j < grid.length; j++) {
                left2Right[i] = Math.max(grid[j][i], left2Right[i]);
            }
        }
        //遍历每个元素，该元素所能增大的最大值为 所处行和所处列 中最小的skyline
        for (int i = 0; i < rowNumber; i++) {
            for (int j = 0; j < columnNumber; j++) {
                int increaseToHeight = Math.min(top2Bottom[j], left2Right[i]);
                res += increaseToHeight - grid[i][j];
            }
        }
        return res;
    }
}