leetcode42 Trapping Rain Water

Posted 2021-03-08 yawenw

 1 """
 2 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.
 3 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!
 4 Example:
 5 Input: [0,1,0,2,1,0,1,3,2,1,2,1]
 6 Output: 6
 7 """
 8 """
 9 每个位置上积水的高度，应该等于min（左边最高的柱子高度，右边最高的柱子高度） -  这个位置上的柱子高度。
10 所以先从左往右遍历把left_max数组生成，left_max[i]代表 height[i] 及其左侧最高的柱子高度。
11 同理生成right_max。
12 然后再遍历一次对于每个位置计算min(left_max[i], right_max[i]) - height[i]就好了
13 """
14 class Solution1:
15     def trap(self, height):
16         max_left = [0] * len(height)
17         max_right = [0] * len(height)
18         res = 0
19         for i in range(len(height)):
20             if i > 0:
21                 max_left[i] = max(max_left[i - 1], height[i])
22             else:
23                 max_left[i] = height[i]
24         for j in range(len(height) - 1, -1, -1):
25             if j < len(height) - 1:
26                 max_right[j] = max(max_right[j + 1], height[j])
27             else:
28                 max_right[j] = height[j]
29         for k in range(len(height)):
30             res += min(max_left[k], max_right[k]) - height[k]
31         return res
32 
33 """
34 自己的想法是创建i和j分别指向每个积水槽的两端
35 积水的体积 = 集水槽的短边*底的长度-中间的每个值
36 但主要卡在 i，j无法寻找
37 我的想法是 height[i]>height[i-1] and height[i]>height[i+1]
38 但对于[2, 0, 2]这种测试案例无法解决
39 """
40 class Solution2:
41     def trap(self, height):
42         n = len(height)
43         i, j = 1, 0
44         res = 0
45         while i < n-1:
46             if height[i] > height[i-1] and height[i] > height[i+1]:
47                 j = i+1
48                 neg = 0
49                 while j < n - 2:
50                     if not (height[j] > height[j - 1] and height[j] > height[j + 1]):
51                         neg += height[j]
52                         j += 1
53                     else:
54                         break
55                 res += min(height[i], height[j]) * (j - i - 1) - neg
56             i = j
57         return res