leetcode42 Trapping Rain Water

"""
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.
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!
Example:
Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
"""
"""
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。
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。
同理生成right_max。
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
"""
class Solution1:
    def trap(self, height):
        max_left = [0] * len(height)
        max_right = [0] * len(height)
        res = 0
        for i in range(len(height)):
            if i > 0:
                max_left[i] = max(max_left[i - 1], height[i])
            else:
                max_left[i] = height[i]
        for j in range(len(height) - 1, -1, -1):
            if j < len(height) - 1:
                max_right[j] = max(max_right[j + 1], height[j])
            else:
                max_right[j] = height[j]
        for k in range(len(height)):
            res += min(max_left[k], max_right[k]) - height[k]
        return res

"""
自己的想法是创建i和j分别指向每个积水槽的两端
积水的体积 = 集水槽的短边*底的长度-中间的每个值
但主要卡在 i，j无法寻找
我的想法是 height[i]>height[i-1] and height[i]>height[i+1]
但对于[2, 0, 2]这种测试案例无法解决
"""
class Solution2:
    def trap(self, height):
        n = len(height)
        i, j = 1, 0
        res = 0
        while i < n-1:
            if height[i] > height[i-1] and height[i] > height[i+1]:
                j = i+1
                neg = 0
                while j < n - 2:
                    if not (height[j] > height[j - 1] and height[j] > height[j + 1]):
                        neg += height[j]
                        j += 1
                    else:
                        break
                res += min(height[i], height[j]) * (j - i - 1) - neg
            i = j
        return res