[Leetcode] 807. Max Increase to Keep City Skyline 解题报告

题目：

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.

思路：

我们定义两个数组rows和cols，分别表示每一行的最大值和每一列的最大值，然后分两步：1）遍历一遍数组，得到rows和cols的值；2）再遍历一遍数组，对于grid[r][c]，如果它小于min(rows[r], cols[c])，那么就可以给它增加min(rows[r], cols[c]) - grid[r][c]，而不影响rows和cols。

算法的时间复杂度是O(mn)，空间复杂度是O(m + n)，其中m和n分别是grid的行个数和列个数。

代码：

class Solution {
public:
    int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
        int row_num = grid.size(), col_num = grid[0].size();
        vector<int> rows(row_num, 0), cols(col_num, 0);
        for (int r = 0; r < row_num; ++r) {
            for (int c = 0; c < col_num; ++c) {
                rows[r] = max(rows[r], grid[r][c]);
                cols[c] = max(cols[c], grid[r][c]);
            }
        }
        int ret = 0;
        for (int r = 0; r < row_num; ++r) {
            for (int c = 0; c < col_num; ++c) {
                int min_value = min(rows[r], cols[c]);
                if (min_value > grid[r][c]) {
                    ret += min_value - grid[r][c];
                }
            }
        }
        return ret;
    }
};