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

class Solution:
# @param A, a list of integers
# @return an integer
def trap(self, A):
# special case
if len(A)<3: return 0 # left most & right most LENGTH=len(A) left\_most = \[0 for i in range(LENGTH)\] right\_most = \[0 for i in range(LENGTH)\] curr\_max = 0 for i in range(LENGTH): if A\[i\] > curr_max:
curr_max = A[i]
left_most[i] = curr_max
curr_max = 0
for i in range(LENGTH-1,-1,-1):
if A[i] > curr_max:
curr_max = A[i]
right_most[i] = curr_max
# sum the trap
sum = 0
for i in range(LENGTH):
sum = sum + max(0,min(left_most[i],right_most[i])-A[i])
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。