标签：inpu not ppi ima ted str http load rgba
Description: Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.
Link: 42. Trapping Rain Water
Examples:
Example 1:
Input: height = [0,1,0,2,1,0,1,3,2,1,2,1] Output: 6 Explanation: The above elevation map (black section) 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. Example 2: Input: height = [4,2,0,3,2,5] Output: 9
思路: 遍历每个位置可储存的雨水，比如[4, 2, 8], 在位置 i = 1, height[i] = 2上, 位置i左边的最高处max_left[i]，右边的最高处max_right[i]，位置i的储存雨水量=min(max_left[i], max_right[i]) - height[i]. 所以我们就用两个数组记录每个位置i的左右边最高处的高度，然后遍历叠加。
class Solution(object): def trap(self, height): """ :type height: List[int] :rtype: int """ if not height: return 0 n = len(height) max_left = [0]*n max_right = [0]*n for i, h in enumerate(height): max_left[i] = max(max_left[i-1], height[i]) max_right[-1] = height[-1] i = n - 2 while i >= 0: max_right[i] = max(max_right[i+1], height[i]) i -= 1 res = 0 for i, h in enumerate(height): res += max(0, (min(max_left[i], max_right[i]) - h)) return res
日期: 2021-04-24 前天和昨天都没有做题，最近学习的热情越来越....马上要deadline了，procrastination
Leetcode** 42. Trapping Rain Water
标签：inpu not ppi ima ted str http load rgba
原文地址：https://www.cnblogs.com/wangyuxia/p/14698021.html
