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;

    }

};