# 接雨水

给定 n 个非负整数表示每个宽度为 1 的柱子的高度图,计算按此排列的柱子,下雨之后能接多少雨水。

 

示例 1:

输入:height = [0,1,0,2,1,0,1,3,2,1,2,1]
输出:
6
解释:
上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图,在这种情况下,可以接 6 个单位的雨水(蓝色部分表示雨水)。

示例 2:

输入:height = [4,2,0,3,2,5]
输出:
9

 

提示:

以下程序实现了这一功能,请你填补空白处内容: ```python class Solution(object): def trap(self, height): ls = len(height) if ls == 0: return 0 res, left = 0, 0 while left < ls and height[left] == 0: left += 1 pos = left + 1 while pos < ls: if height[pos] >= height[left]: res += self.rain_water(height, left, pos) left = pos pos += 1 elif pos == ls - 1: max_value, max_index = 0, pos for index in range(left + 1, ls): if height[index] > max_value: max_value = height[index] max_index = index res += self.rain_water(height, left, max_index) left = max_index pos = left + 1 else: pos += 1 return res def rain_water(self, height, start, end): if end - start <= 1: return 0 min_m = min(height[start], height[end]) _________________________ step = 0 for index in range(start + 1, end): if height[index] > 0: step += height[index] return res - step if __name__ == '__main__': s = Solution() print (s.trap([2,6,3,8,2,7,2,5,0])) ``` ## template ```python class Solution(object): def trap(self, height): ls = len(height) if ls == 0: return 0 res, left = 0, 0 while left < ls and height[left] == 0: left += 1 pos = left + 1 while pos < ls: if height[pos] >= height[left]: res += self.rain_water(height, left, pos) left = pos pos += 1 elif pos == ls - 1: max_value, max_index = 0, pos for index in range(left + 1, ls): if height[index] > max_value: max_value = height[index] max_index = index res += self.rain_water(height, left, max_index) left = max_index pos = left + 1 else: pos += 1 return res def rain_water(self, height, start, end): if end - start <= 1: return 0 min_m = min(height[start], height[end]) res = min_m * (end - start - 1) step = 0 for index in range(start + 1, end): if height[index] > 0: step += height[index] return res - step if __name__ == '__main__': s = Solution() print (s.trap([2,6,3,8,2,7,2,5,0])) ``` ## 答案 ```python res = min_m * (end - start - 1) ``` ## 选项 ### A ```python res = min_m * (end - start) / 2 ``` ### B ```python res = min_m * (end + start) / 2 ``` ### C ```python res = min_m * (end - start) ```