[LeetCode]題解（python）：042-Trapping Rain Water

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題目來源

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.

題意分析

Input: the height of the bars as list

Output: the volumn of the water the bars can save

Conditions:注意到因為bar的高度不一樣，所以中間的bar可以容納一定量的水，題意就是求容納的水的大小

題目思路

注意到每一個bar所能容納的水的量，就是其左右最高的bar中的較小值減去該bar的高度，所以先建立一個leftMostHigh的list記錄每一個bar之前的最高bar值，然後從後往前遍曆，一邊更新右邊最高的bar值，同時計算每一個bar所能容納的水量

AC代碼（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 = 013         for i in range(l):14             leftMostHigh[i] = leftmax15             if height[i] > leftmax:16                 leftmax = height[i]17 18         rightmax = 019         sum = 020         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 sum27 28 29 height = [0,1,0,2,1,0,1,3,2,1,2,1]30 s = Solution()31 print(s.trap(height))

 

[LeetCode]題解（python）：042-Trapping Rain Water