[LintCode 363] 接雨水(Python)

题目描述

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.

样例

如上图所示，海拔分别为 [0,1,0,2,1,0,1,3,2,1,2,1], 返回 6.

思路

观察图片可知，接满雨水后，总体呈现先增后降的趋势。所以先遍历一遍，找到峰值的下标。再从两头向峰值遍历，不符合递增趋势的数值，都是需要计算填充差值的。

代码

class Solution:
"""
@param: heights: a list of integers
@return: a integer
"""
def trapRainWater(self, heights):
# write your code here
if heights is None or len(heights) <= 2:
return 0
max_id = 0
for i in range(1, len(heights)):
if heights[i] > heights[max_id]:
max_id = i

res = 0
s = 0
cur = heights[s]
while s < max_id:
if heights[s] >= cur:
cur = heights[s]
else:
res += 1 * (cur - heights[s])
s += 1

e = len(heights) - 1
cur = heights[e]
while e > max_id:
if heights[e] >= cur:
cur = heights[e]
else:
res += 1 * (cur - heights[e])
e -= 1
return res

复杂度分析

时间复杂度O(n)，空间复杂度O(1)。