LeetCode刷题（10）

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.

class Solution(object):
    def trap(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        l, r  = 0, len(height) - 1
        water = 0
        while l < r and height[l+1] >= height[l]:
            l += 1
        while l < r and height[r-1] >= height[r]:
            r -= 1
        while(l < r):
            if height[l] < height[r]:
                level = height[l]
                l += 1
                while l < r and height[l] <= level:
                    water = water + level - height[l]
                    l += 1
            else:
                level = height[r]
                r -= 1
                while l < r and height[r] <= level:
                    water = water + level - height[r]
                    r -= 1
        return water 

从地势的两头算起，如果中间高两头低是无法存水的。当存在中间的洼地时，存在存水的空间，并以此时的两端作为边沿，以两端较低的一个作为水平，往中间测算，如果靠近中间的地势低于水平，则存水量应为水平减去地势，如果高于水平，则该点作为新的边沿，重新计算水平。