原題地址：https://oj.leetcode.com/problems/trapping-rain-water/

題意：

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.

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 rain water (blue section) are being trapped.?Thanks Marcos?for contributing this image!

解題思路：模擬法。開辟一個數組leftmosthigh，leftmosthigh[i]為A[i]之前的最高的bar值，然后從后面開始遍歷，用rightmax來記錄從后向前遍歷遇到的最大bar值，那么min(leftmosthigh[i], rightmax)-A[i]就是在第i個bar可以儲存的水量。例如當i=9時，此時leftmosthigh[9]=3,而rightmax=2，則儲水量為2-1=1，依次類推即可。這種方法還是很巧妙的。時間復雜度為O(N)。

代碼：

class Solution:# @param A, a list of integers# @return an integerdef trap(self, A):leftmosthigh = [0 for i in range(len(A))]leftmax = 0for i in range(len(A)):if A[i] > leftmax: leftmax = A[i]leftmosthigh[i] = leftmaxsum = 0rightmax = 0for i in reversed(range(len(A))):if A[i] > rightmax: rightmax = A[i]if min(rightmax, leftmosthigh[i]) > A[i]:sum += min(rightmax, leftmosthigh[i]) - A[i]return sum

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