Leetcode||42. 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.

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 Marcosfor contributing this image!

模拟法。开辟一个数组leftmostheigh，leftmostheigh[i]为A[i]之前的最高的bar值，然后从后面开始遍历，用rightmax来记录从后向前遍历遇到的最大bar值，那么min(leftmostheigh[i], rightmax)-A[i]就是在第i个bar可以储存的水量。

class Solution(object):    def trap(self, height):        """        :type height: List[int]        :rtype: int        """        leftmostheigh = [0 for i in range(len(height))]        leftmax = 0        for i in range(len(height)):        if height[i] > leftmax:        leftmax = height[i]        leftmostheigh[i] = leftmax        sum = 0        rightmax = 0        for i in reversed(range(len(height))):            if height[i] > rightmax:            rightmax = height[i]            if min(rightmax, leftmostheigh[i]) > height[i]:                sum += min(rightmax, leftmostheigh[i]) - height[i]        return sum