leetcode#42 Trapping rain water 这道题十分有意思,可以用很多方法做出来,每种方法的思想都值得让人细细体会. 42. Trapping Rain WaterGiven 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 ex

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

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

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

Container With Most Water 题目链接 Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which togeth

42. Trapping Rain Water Problem's Link ---------------------------------------------------------------------------- Mean: 在坐标上给你一些竖直放置的条形积木,问你这个积木能够容纳多少液体. analyse: 首先找出最高的积木,然后从前往后一直扫到最高积木,从后往前一直扫到最高积木,两部分体积相加即可. Time complexity: O(N) view code ;  

LeetCode传送门 https://leetcode.com/problems/trapping-rain-water/ 目标:找出积木能容纳的水的“面积”,如图中黑色部分是积木,蓝色为可容纳水的部分 假设:积木宽度均为1 输入:各个积木的高度 输出:所有积木能容纳水的“面积” 思考过程 1. 逐一求积木的间隔似乎不太容易.特别对于图中3-7积木间的容积,如果可以先求底部(4-6间)的容积,当求解上层(3-7)的容积时,还需要做额外的处理,如减掉底部的高度. 2. 既然如此,可否先求出3-7

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 42: Trapping Rain Waterhttps://oj.leetcode.com/problems/trapping-rain-water/ Given n non-negative integers representing an elevation map where the width of each bar is 1,compute how much wa

leetcode - 42. Trapping Rain Water - Hard descrition 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],

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

https://oj.leetcode.com/problems/trapping-rain-water/ Trapping Rain WaterGiven 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,

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 ma

这题放上来是因为自己第一回见到这种题,觉得它好玩儿 =) 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], retu

LeetCode 42. Trapping Rain Water Python解法 解题思路: 本思路需找到最高点左右遍历,时间复杂度O(nlogn),以下为向左遍历的过程. 将每一个点的高度和索引存成一个元组 (val, idx) 找到最高的点(可能有多个,任取一个),记为 (now_val, now_idx). 向左找第一个val不大于now_val的点(left_val, left_idx). 以left_val作为水平面,用height[left_val+1, now_val-1]中每一

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. The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of

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

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 tha

11. Container With Most Water https://www.cnblogs.com/grandyang/p/4455109.html 用双指针向中间滑动,较小的高度就作为当前情况的高度,然后循环找容量的最大值. 不管两个指针中间有多少高度的柱子,只管两头,因为两头的才决定最大容量. class Solution { public: int maxArea(vector<int>& height) { if(height.empty()) ; ; ,end = h

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. The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of

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