博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
807. Max Increase to Keep City Skyline
阅读量:6787 次
发布时间:2019-06-26

本文共 2552 字,大约阅读时间需要 8 分钟。

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: 35Explanation: 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.

解题思路

先得到每一行所对应最高建筑的高度row[i]和每一列对应的最高建筑的高度col[j]。要增加建筑面积,但是为了不改变横向和纵向看去城市最高的建筑高度,

对于每一个建筑[i][j]:

  (1)是其所在行或列的最高建筑时,不能改变其高度。

  (2)其不是所在行或列的最高建筑时,可以增加高度至col[j]和row[i]中最小值。

代码:

1 int maxIncreaseKeepingSkyline(int** grid, int gridRowSize, int *gridColSizes) { 2     int* row = (int*) malloc(sizeof(int) * gridRowSize); 3     int* col = (int*) malloc(sizeof(int) * gridColSizes[0]); 4     for (int i = 0; i < gridRowSize; ++i) { 5         int max = 0; 6         for (int j = 0; j < gridColSizes[0]; ++j) { 7             if (grid[i][j] > max) 8                 max = grid[i][j]; 9         }10         row[i] = max;11     }12     for (int i = 0; i < gridColSizes[0]; ++i) {13         int max = 0;14         for (int j = 0; j < gridRowSize; ++j) {15             if (grid[j][i] > max)16                 max = grid[j][i];17         }18         col[i] = max;19     }20     int added = 0;21     for (int i = 0; i < gridRowSize; ++i) {22         for (int j = 0; j < gridColSizes[0]; ++j) {23             int max = row[i] < col[j] ? row[i] : col[j];24             if (grid[i][j] < max)25                 added += max - grid[i][j];26         }27     }28     return added;29 }

 

转载于:https://www.cnblogs.com/gsz-/p/9379704.html

你可能感兴趣的文章
2003应用池假死常见问题和解决方法
查看>>
使用javascript的日期函数
查看>>
c# : use xsd 校验 xml
查看>>
mybatis初接触
查看>>
没有测试的开发是多么的悲催哇
查看>>
awk的日志模块追加日期时间字段的方案
查看>>
[转]高级SQL注入:混淆和绕过
查看>>
System.IO.Path 文件名、路径、扩展名处理
查看>>
类的成员修饰符
查看>>
课堂训练
查看>>
HDU 5464:Clarke and problem
查看>>
Web服务器禁止range请求
查看>>
php编译GD库 JPEG Support
查看>>
【转】着色中的数学和物理原理
查看>>
overflow的使用
查看>>
Position Independent Code (PIC) in shared libraries on x64
查看>>
CNBLOG上几位.NET大牛的博客地址(转)
查看>>
接口继承和实现继承的区别
查看>>
spring 的自建request请求
查看>>
数组的相关知识
查看>>