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

---

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）

---

 _author_ = "YE"
 # -*- coding:utf-8 -*-

 class Solution(object):
     def trap(self, height):
         """
         :type height: List[int]
         :rtype: int
         """
         l = len(height)
         leftMostHigh = [0 for i in range(len(height))]
         leftmax = 0
         for i in range(l):
             leftMostHigh[i] = leftmax
             if height[i] > leftmax:
                 leftmax = height[i]

         rightmax = 0
         sum = 0
         for i in reversed(range(l)):
             if min(rightmax, leftMostHigh[i]) > height[i]:
                 sum = sum + min(rightmax, leftMostHigh[i]) - height[i]
             if height[i] > rightmax:
                 rightmax = height[i]

         return sum

 height = [0,1,0,2,1,0,1,3,2,1,2,1]
 s = Solution()
 print(s.trap(height))