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!

class Solution {
public:
    /**
     * @param heights: a vector of integers
     * @return: a integer
     */
    int trapRainWater(vector<int> &heights) {
        // write your code here
        int n = heights.size();
        if (n < 1)
        {
            return 0;
        }
        int result = 0;
        int rightMax[n];
        rightMax[n-1] = heights[n-1];
        for (int i = n-2; i >= 0; i--)
        {
            if (heights[i] > rightMax[i+1])
            {
                rightMax[i] = heights[i];
            }
            else
            {
                rightMax[i] = rightMax[i+1];
            }
        }

        int begin = 0;
        while (begin < n && heights[begin] == 0)
        {
            begin++;
        }

        for (int i = begin+1; i < n; i++)
        {
            if (heights[i] >= heights[begin] || heights[i] == rightMax[i])
            {
                int top = min(heights[i], heights[begin]);
                for (int j = begin+1; j < i; j++)
                {
                    if (heights[j] < top)
                    {
                        result += (top - heights[j]);
                    }
                }

                begin = i;
            }
        }

        return result;
    }
};