LeetCode #807 - Max Increase to Keep City Skyline

最新推荐文章于 2022-01-12 14:41:30 发布

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本文探讨了一个有趣的问题：在一个二维网格中，如何增加建筑的高度，使得从四个方向看的天际线保持不变，同时求出可以增加的最大总高度。通过找出每一行和每一列的最大值，我们可以确定新的网格形状，确保天际线的一致性。

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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.

class Solution {
public:
    int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
        int n=grid.size();
        vector<int> max_col(n,0);
        vector<int> max_row(n,0);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                max_row[i]=max(max_row[i],grid[i][j]);
                max_col[j]=max(max_col[j],grid[i][j]);
            }
        }
        
        int result=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                result+=min(max_row[i],max_col[j])-grid[i][j];
        return result;
    }
};