leetcode 【 Trapping Rain Water 】python 實現

標籤：

題目：

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!

 

代碼：oj測試通過 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 = 013         for i in range(LENGTH):14             if A[i] > curr_max:15                 curr_max = A[i]16             left_most[i] = curr_max17         curr_max = 018         for i in range(LENGTH-1,-1,-1):19             if A[i] > curr_max:20                 curr_max = A[i]21             right_most[i] = curr_max22         # sum the trap23         sum = 024         for i in range(LENGTH):25             sum = sum + max(0,min(left_most[i],right_most[i])-A[i])26         return sum

 

思路：

一句話：某個Position能放多少水，取決於左右兩邊最小的有這個Position的位置高。

可以想象一下實體環境，一個位置要能存住水，就得保證這個Position處於一個低洼的位置。怎麼才能滿足低洼位置的條件呢？左右兩邊都得有比這個position高的元素。如果才能保證左右兩邊都有比這個position高的元素存在呢？只要左右兩邊的最大值中較小的一個比這個Position大就可以了。

leetcode 【 Trapping Rain Water 】python 實現