807. Max Increase to Keep City Skyline （Python）

最新推荐文章于 2024-08-16 09:00:20 发布

Karen_Yu_

最新推荐文章于 2024-08-16 09:00:20 发布

阅读量670

点赞数

分类专栏： LeetCode python 文章标签： python

本文链接：https://blog.youkuaiyun.com/Karen_Yu_/article/details/122228063

版权

python 同时被 2 个专栏收录

25 篇文章

订阅专栏

LeetCode

2 篇文章

订阅专栏

给定一个n×n的矩阵表示城市中建筑物的高度，任务是计算在不改变任何方向天际线的前提下，能将所有建筑物高度增加的最大总和。例子展示了如何通过选取行和列的最大值来确定可增加的高度，并给出了一个解决方案，但效率较低。文章探讨了不同的解题思路和代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Medium

There is a city composed of n x n blocks, where each block contains a single building shaped like a vertical square prism. You are given a 0-indexed n x n integer matrix grid where grid[r][c] represents the height of the building located in the block at row r and column c.

A city's skyline is the the outer contour formed by all the building when viewing the side of the city from a distance. The skyline from each cardinal direction north, east, south, and west may be different.

We are allowed to increase the height of any number of buildings by any amount (the amount can be different per building). The height of a 0-height building can also be increased. However, increasing the height of a building should not affect the city's skyline from any cardinal direction.

Return the maximum total sum that the height of the buildings can be increased by without changing the city's skyline from any cardinal direction.

Example 1:

Input: grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
Output: 35
Explanation: The building heights are shown in the center of the above image.
The skylines when viewed from each cardinal direction are drawn in red.
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] ]

Example 2:

Input: grid = [[0,0,0],[0,0,0],[0,0,0]]
Output: 0
Explanation: Increasing the height of any building will result in the skyline changing.

Constraints:

n == grid.length
n == grid[r].length
2 <= n <= 50
0 <= grid[r][c] <= 100

完全不用动脑子版：

class Solution(object):
    def maxIncreaseKeepingSkyline(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        #理论上就是扫描整个矩阵，选取该点所在行的最大值和所在列的最大值之中的最小值
        #得到原值与新值的差值
        rowlen = len(grid)
        collen = len(grid[0])
        rowMax = []
        colMax = []
        
        #先得到每行最大值的数组
        for i in range(rowlen):
            rowmax = max(grid[i])
            rowMax.append(rowmax)
        
        newGrid = copy.deepcopy(grid)
        
        #按照之前48. Rotate Image方法翻转矩阵
        #得到每列最大值的数组
        for i in range(rowlen/2):
            for j in range(collen):
                newGrid[i][j],newGrid[collen-1-i][j]=newGrid[collen-1-i][j],newGrid[i][j]
        for i in range(rowlen):
            for j in range(i,collen):
                newGrid[i][j],newGrid[j][i]=newGrid[j][i],newGrid[i][j]
        
        for i in range(rowlen):
            colmax = max(newGrid[i])
            colMax.append(colmax)
        #每个位置对应计算增加的高度
        IncSky = 0
        for i in range(rowlen):
            for j in range(collen):
                newGrid[i][j]=min(rowMax[i],colMax[j])
                IncSky += newGrid[i][j]-grid[i][j]
                
        return IncSky

不过效率感人

提到的48题：https://leetcode.com/problems/rotate-image/

看了一下逻辑上好像和别的差距不是很大的样子：

https://www.cnblogs.com/fuxuemingzhu/p/15436313.html

不过类似这种写法我好像一直不太习惯orz：

for i in range(M):
        rows[i] = max(grid[i][j] for j in range(N))
for j in range(N):
        cols[j] = max(grid[i][j] for i in range(M))

https://blog.youkuaiyun.com/danspace1/article/details/87789959

https://www.jianshu.com/p/36b8b74bf489

似乎大部分都是暴力解的样子，本打表+暴力爱好者狂喜