42. 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

解题要点：

用Dynamic Programming，先从左至右搜索一遍，记录最大值，再从右至左搜索一遍，同样记录最大值。最后遍历的时候取min(left[i], right[i])，再减去height[i]，记录为结果。

class Solution(object):
    def trap(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        if height == None or len(height) == 0:
            return 0
        leftDP = [0] * len(height)
        rightDP = [0] * len(height)
        leftDP[0] = height[0]
        rightDP[-1] = height[-1]
        for i in range(1, len(height)):
            leftDP[i] = max(leftDP[i-1], height[i])
        for j in range(len(height)-2, -1, -1):
            rightDP[j] = max(rightDP[j+1], height[j])
        res = 0
        for k in range(len(height)):
            res += min(leftDP[k], rightDP[k]) - height[k]
        return res