leetcode-807-Max-Increase-to-Keep-City-Skyline

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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?

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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.

分析

题目中给出一个二维数组，每个元素表示一栋建筑的高度，从东南西北四个方向看过去能看到城市的天际线，现在问不改变天际线的情况下，城市还能增加多少高度。

一个建筑物要增加高度且不改变天际线，那么它增加高度后不应该超过它所在行或列的最大值，也就是说所在行或列的最大值中的较小者，可以先遍历一遍，用两个一维数组记录，在遍历以便进行累加即可。时间复杂度是 $O(N^2)$，空间复杂度是 $O(N)$。

解决方案1(Java)

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class Solution {
    public int maxIncreaseKeepingSkyline(int[][] grid) {
        int N = grid.length;
        int[] rowMax = new int[N];
        int[] colMax = new int[N];
        
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                rowMax[i] = Math.max(rowMax[i], grid[i][j]);
                colMax[j] = Math.max(colMax[j], grid[i][j]);
            }
        }
        
        int result = 0;
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                result += Math.min(rowMax[i], colMax[j]) - grid[i][j];
            }
        }
        return result;
    }
}

题目来源