leetcode-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?

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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.

这一题是计算最大的建筑容量，具体解题思路为先取出每一行每一列的最大值 top, left。 然后新构成的矩阵的每一个值 new[i][j]为top[j]和left[i]中较小的值。最后计算两个矩阵中所有元素的差即可。具体解法如下。

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class Solution {
public:
    int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
        int sum_p = 0, sum_a = 0;
        vector<int> top,left;

        top = grid[0];

        for(int i = 0; i < grid.size(); i++)
        {
            left.push_back(grid[i][0]);
            for(int j = 0; j < grid[0].size(); j++)
            {
                if(top[j] < grid[i][j])
                {
                    top[j] = grid[i][j];
                }

                if(left[i] < grid[i][j])
                {
                    left[i] = grid[i][j];
                }
                sum_p += grid[i][j];
            }
        }

        for(int i = 0; i < grid.size(); i++)
        {
            int temp_t = left[i];
            for(int j = 0; j < grid[0].size(); j++)
            {
                sum_a += (temp_t < top[j]?temp_t:top[j]);
            }
        }

        return (sum_a-sum_p);
    }
};