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题目来源
https://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 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.
题意分析
Input: the height of the bars as list
Output: the volumn of the water the bars can save
Conditions:注意到因为bar的高度不一样，所以中间的bar可以容纳一定量的水，题意就是求容纳的水的大小
题目思路
注意到每一个bar所能容纳的水的量，就是其左右最高的bar中的较小值减去该bar的高度，所以先建立一个leftMostHigh的list记录每一个bar之前的最高bar值，然后从后往前遍历，一边更新右边最高的bar值，同时计算每一个bar所能容纳的水量
AC代码（Python）
1 _author_ = "YE" 2 # -*- coding:utf-8 -*- 3 4 class Solution(object): 5 def trap(self, height): 6 """ 7 :type height: List[int] 8 :rtype: int 9 """ 10 l = len(height) 11 leftMostHigh = [0 for i in range(len(height))] 12 leftmax = 0 13 for i in range(l): 14 leftMostHigh[i] = leftmax 15 if height[i] > leftmax: 16 leftmax = height[i] 17 18 rightmax = 0 19 sum = 0 20 for i in reversed(range(l)): 21 if min(rightmax, leftMostHigh[i]) > height[i]: 22 sum = sum + min(rightmax, leftMostHigh[i]) - height[i] 23 if height[i] > rightmax: 24 rightmax = height[i] 25 26 return sum 27 28 29 height = [0,1,0,2,1,0,1,3,2,1,2,1] 30 s = Solution() 31 print(s.trap(height))
[LeetCode]题解（python）：042-Trapping Rain Water
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原文地址：http://www.cnblogs.com/loadofleaf/p/5042960.html
