42. Trapping Rain Water leetcode Python 2016 new season

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本文介绍了一种算法，用于计算给定高度数组所代表的地形中，在雨后能够储存的雨水量。通过分析数组并计算每个点两侧的最大高度，可以得出最终能够储存的雨水总量。

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.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

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!

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class Solution(object):
    def trap(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        if not height:
            return 0
        height_len = len(height)
        maxLeft = [0 for _ in range(height_len)]
        maxRight = [0 for _ in range(height_len)]
        max_value = height[0]
        for index in range(height_len):
            maxLeft[index] = max_value
            if max_value < height[index]:
                max_value = height[index]
        max_value = height[height_len - 1]
        for index in reversed(range(height_len)):
            maxRight[index] = max_value
            if max_value < height[index]:
                max_value = height[index]
        result = 0
        for index in range(height_len):
            water = min(maxRight[index], maxLeft[index]) - height[index]
            if water > 0:
                result += water
        return result