LeetCode | Trapping Rain Water

题目：

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!

思路：

首先从两边向中间找到最高的那一根柱子。

然后从两边分别向中间移动并灌水。我们用floor来表示当前水平面，如果水平面比柱子高，那么这个区域可以灌水；如果水平面比柱子低，那么向里的水平面可以增加。

代码：

class Solution {public:    int trap(int A[], int n) {        if(n<2)            return 0;        int highest=A[0];        int hindex = 0;        for(int i=1;i<n;i++)        {            if(A[i]>highest)            {                highest = A[i];                hindex = i;            }        }                int water=0;                int floor=0;;        for(int i=0;i<hindex;i++)        {            if(floor>=A[i])            {                water+=floor-A[i];            }            else            {                floor=A[i];            }        }                floor=0;;        for(int i=n-1;i>hindex;i--)        {            if(floor>=A[i])            {                water+=floor-A[i];            }            else            {                floor=A[i];            }        }                return water;    }};