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…

题目:接雨水 难度:hard 题目内容: 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…

题意: 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. watermark/2/text/aHR0cDovL2Jsb2cuY3Nkb…

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…

# -*- 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 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…

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],…

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…

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…