LeetCode重点题系列之【42. Trapping Rain Water】

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本文介绍了一种计算雨后地面积水的方法。通过两次遍历高度数组，分别从左到右和从右到左记录最大高度，并据此计算每个位置能够容纳的雨水量。最终返回所有位置的积水总量。

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.

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

This problem is very hard. First, you should scan from left to right, keep track current largest height. Second, scan for the right to left, keep tracking the current largest height. Maintain a result list, use ith element in the minimal height of left to right list and right to the left list to minus ith element in height. Return sum of res.

class Solution(object):
    def trap(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        if len(height)<3:
            return 0
        left2right=[0 for _ in range(len(height))]
        right2left=[0 for _ in range(len(height))]
        left2right[0]=height[0]
        right2left[len(height) - 1] = height[len(height) - 1]
        for i in range(1,len(height)):
            if height[i]>left2right[i-1]:
                left2right[i]=height[i]
            else:
                left2right[i]=left2right[i-1]
        for i in range(len(height)-2,-1,-1):
            if height[i]>right2left[i+1]:
                right2left[i]=height[i]
            else:
                right2left[i]=right2left[i+1]
        res=[]
        for i,item in enumerate(height):
            res.append(min(left2right[i],right2left[i])-item)
        return sum(res)