[LeetCode] Max Increase to Keep City Skyline

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题目描述

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.
二维数组存储楼房的高度，从左到右或从上到下都有最大值。要求对于每个节点尽可能升高高度，要求升高之后之前从左到右以及从上到下的最大值依然最大。要求返回最高的升高高度。
思路是遍历二维数组，求出从左到右以及从上到下的最高楼房高度，然后在遍历一遍二维数组中的所有值，当前可以到达的最高高度是当前位置从左到右和从上到下的较小值，不断累加得到最后结果。

C++ 实现

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class Solution {
public:
    int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
        int row_size = grid.size();
        if (row_size == 0) return 0;
        int col_size = grid[0].size();
        vector<int> top_2_bottom(row_size);
        vector<int> left_2_right(col_size);
        for (int i = 0; i < row_size; i++)
        {
            top_2_bottom[i] = *max_element(grid[i].begin(), grid[i].end());
        }
        for (int j = 0; j < col_size; j++)
        {
            int max_element = grid[0][j];
            for (int i = 1; i < row_size; i++)
            {
                max_element = max(max_element, grid[i][j]);
            }
            left_2_right[j] = max_element;
        }
        int res = 0;
        for (int i = 0; i < row_size; i++)
        {
            for (int j = 0; j < col_size; j++)
            {
                res += min(top_2_bottom[i], left_2_right[j]) - grid[i][j];
            }
        }
        return res;
    }
};
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