[LintCode] Trapping Rain Water 收集雨水

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

Have you met this question in a real interview?

Yes

Example

Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

Challenge

O(n) time and O(1) memory

O(n) time and O(n) memory is also acceptable.

s

时间： 2024-12-14 03:12:19

[LintCode] Trapping Rain Water 收集雨水的相关文章

[LeetCode]101. Trapping Rain Water收集雨水

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. The above elevation map is represented by a

[poj] The Wedding Juicer | [lintcode] Trapping Rain Water II

问题描述 给定一个二维矩阵,每个元素都有一个正整数值,表示高度.这样构成了一个二维的.有高度的物体.请问该矩阵可以盛放多少水? 相关题目:POJ The Wedding Juicer Description Farmer John's cows have taken a side job designing interesting punch-bowl designs. The designs are created as follows: * A flat board of size W cm

[LintCode] Trapping Rain Water II

Trapping Rain Water II Given n x m non-negative integers representing an elevation map 2d where the area of each cell is 1 x 1, compute how much water it is able to trap after raining. Example Given 5*4 matrix [12,13,0,12] [13,4,13,12] [13,8,10,12] [

[LeetCode] Trapping Rain Water II 收集雨水之二

Given an m x n matrix of positive integers representing the height of each unit cell in a 2D elevation map, compute the volume of water it is able to trap after raining. Note: Both m and n are less than 110. The height of each unit cell is greater th

leetCode 42.Trapping Rain Water(凹槽的雨水) 解题思路和方法

Trapping Rain Water Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. The above elevation map

[LeetCode][JavaScript]Trapping Rain Water

Trapping Rain Water Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. The above elevation map

每日算法之三十三：Trapping Rain Water

这是一个很有意思的问题,求解最大容积问题,值得动脑筋想一想. 原题如下: Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. The ab

No.42 Trapping Rain Water

1 class Solution 2 { 3 public: 4 int trap(vector<int> &height) 5 {/* 6 雨水覆盖问题: 7 每一个bar能承受的水量为:它左侧所有中最高的和右侧所有中最高的中取最小值作为一个瓶颈[否则也留不住], 8 若该值大于当前bar的高度,其差值即为所求 9 累加所有bar能承受的水量即为所求 10 法三:[左右两个指针] 11 找到最高的位置,将数组分为两部分: 12 对左侧数据,因为右边已经有了最高值,所以,类似法一,找到

刷题42. Trapping Rain Water

一.题目说明 题目是42. Trapping Rain Water,翻译起来就是"接雨水".给n个非负正数代表高度,每个正数宽度为1,让计算能多少雨水.题目难度是Hard 二.我的解法 这个题目是找"坑",然后计算里面可以存的"雨".总共提交了5次,前面几次都是边界错误. 代码如下: #include<iostream> #include<vector> using namespace std; class Solutio