zoukankan      html  css  js  c++  java
  • 807. Max Increase to Keep City Skyline

    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] ]
    
    class Solution:
        def maxIncreaseKeepingSkyline(self, grid):
            """
            :type grid: List[List[int]]
            :rtype: int
            """
            a = []
            b = []
            for i in range(len(grid)):
                max = 0
                for j in range(len(grid[0])):
                    if grid[i][j]>max:
                        max = grid[i][j]
                a.append(max)
            for j in range(len(grid[0])):
                max = 0
                for i in range(len(grid)):
                    if grid[i][j]>max:
                        max = grid[i][j]
                b.append(max)
            sum = 0
            for i in range(len(grid)):
                for j in range(len(grid[0])):
                    sum += min(a[i],b[j]) - grid[i][j]
            return sum
    

    例子讲得很清楚,按照例子里的思路写就可以了。

  • 相关阅读:
    【JVM基础】JVM垃圾回收机制算法
    【java基础】- java双亲委派机制
    Java基础(一)
    JVM
    冷知识: 不会出现OutOfMemoryError的内存区域
    疯狂Java:突破程序员基本功的16课-李刚编著 学习笔记(未完待续)
    nor flash之写保护
    spinor/spinand flash之高频通信延迟采样
    nor flash之频率限制
    使用littlefs-fuse在PC端调试littlefs文件系统
  • 原文地址:https://www.cnblogs.com/bernieloveslife/p/9700416.html
Copyright © 2011-2022 走看看