[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)。