leetcode 【 Trapping Rain Water 】python 实现
题目：
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!
代码：oj测试通过 Runtime: 91 ms
1 class Solution: 2 # @param A, a list of integers 3 # @return an integer 4 def trap(self, A): 5 # special case 6 if len(A)<3: 7 return 0 8 # left most & right most 9 LENGTH=len(A) 10 left_most = [0 for i in range(LENGTH)] 11 right_most = [0 for i in range(LENGTH)] 12 curr_max = 0 13 for i in range(LENGTH): 14 if A[i] > curr_max: 15 curr_max = A[i] 16 left_most[i] = curr_max 17 curr_max = 0 18 for i in range(LENGTH-1,-1,-1): 19 if A[i] > curr_max: 20 curr_max = A[i] 21 right_most[i] = curr_max 22 # sum the trap 23 sum = 0 24 for i in range(LENGTH): 25 sum = sum + max(0,min(left_most[i],right_most[i])-A[i]) 26 return sum
思路：
一句话：某个Position能放多少水，取决于左右两边最小的有这个Position的位置高。
可以想象一下物理环境，一个位置要能存住水，就得保证这个Position处于一个低洼的位置。怎么才能满足低洼位置的条件呢？左右两边都得有比这个position高的元素。如何才能保证左右两边都有比这个position高的元素存在呢？只要左右两边的最大值中较小的一个比这个Position大就可以了。
