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.

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：这道题先找出中间最高的柱子，记下位置maxIndex。然后分别从两边开始向中间循环遍历。先从左边开始记下前一个的柱子高度sum1，然后与后一个的柱子高度相比较，如果比后面的高，则他们两的差额累加到water，反之，更新sum1。再者，从右边开始，相同思路。这样就可以计算出最后储水量。

class Solution {
public:
    int trap(int A[], int n) {
        if(n<2)
            return 0;
        int max=A[0];
        int maxIndex=0;
        for(int i=1;i<n;i++)
        {
            if(A[i]>max)
            {
                max=A[i];
                maxIndex=i;
            }
        }
        int water=0;
        int sum1=0;
        for(int i=0;i<maxIndex;i++)
        {
            if(sum1>A[i])
            {
                water+=sum1-A[i];
            }
            else
                sum1=A[i];
        }
        int sum2=0;
        for(int i=n-1;i>maxIndex;i--)
        {
            if(sum2>A[i])
            {
                water+=sum2-A[i];
            }
            else
                sum2=A[i];
        }
        return water;
    }
};