Trapping Rain Water 接雨水 python

 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.

思路分析：

找到最高柱子，从两端逼近，只要是递增不可能存雨水，需要一个peak移动着时刻记录移动中遇到的临时最高的柱子。时间复杂度O(n)

代码如下：

# coding=utf-8

"""接雨水"""


class Solution(object):
    def trap_rainwater(self, listp):

        movepeak = 0
        triprain = 0
        maxindex = 0
        for i in range(1, len(listp)):
            if listp[i] > listp[maxindex]:
                maxindex = i
        for i in range(0, maxindex):
            if movepeak < listp[i]:
                movepeak = listp[i]
            else:
                triprain += movepeak - listp[i]
        movepeak = 0

        for j in range(len(listp) - 1, maxindex, -1):
            if movepeak < listp[j]:
                movepeak = listp[j]
            else:
                triprain += movepeak - listp[j]
        return triprain

if __name__ == '__main__':
    s = Solution()
    listp = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]
    res = s.trap_rainwater(listp)
    print res