Problem #30 [Medium]

This problem was asked by Facebook.

You are given an array of non-negative integers that represents a two-dimensional elevation map where each element is unit-width wall and the integer is the height. Suppose it will rain and all spots between two walls get filled up.

Compute how many units of water remain trapped on the map in O(N) time and O(1) space.

For example, given the input [2, 1, 2], we can hold 1 unit of water in the middle.

Given the input [3, 0, 1, 3, 0, 5], we can hold 3 units in the first index, 2 in the second, and 3 in the fourth index (we cannot hold 5 since it would run off to the left), so we can trap 8 units of water.
接雨水问题。对于每一个高度i，我们要找的是它两边的大值的最小值。
首先，很容易想到从左到右遍历，从右到左遍历分别找到比当前高度大的值，然后拿到最小值减去当前高度就是可以接到的雨水量。
空间复杂度为O(N)， 时间复杂度为O(N), N为height数组的长度。

def trap(height):
        n = len(height)
        pre_max = [-1] *n
        suf_max = [-1] * n
        pre_max[0] = height[0]
        suf_max[-1] = height[-1]
        for i in range(1, n):
            pre_max[i] = max(pre_max[i-1], height[i])
        for i in range(n-2, -1, -1):
            suf_max[i] = max(suf_max[i+1], height[i])
        
        ans = 0
        for h, pre, suf in zip(height, pre_max, suf_max):
             ans += min(pre,suf) - h
        return ans
        

height = [3,0,1,3,0,5]
print(trap(height))

接下来思考如何才能做到O(1)的空间复杂度呢？
那就是单调栈。
单调栈的总结见我的这篇博客