LeetCode: 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:
    int trap(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (A == NULL || n == 0) return 0;
        vector left(n);
        vector right(n);
        
        int leftMax = 0;
        int rightMax = 0;
        for (int i = 0, j = n-1; i < n; ++i, --j)
        {
            if (A[i] > leftMax)
                leftMax = A[i];
            left[i] = leftMax;
            
            if (A[j] > rightMax)
                rightMax = A[j];
            right[j] = rightMax;
        }
        
        int max = 0;
        int sum = 0;
        for (int i = 0; i < n; ++i)
        {
            max = left[i] < right[i] ? left[i] : right[i];
            sum += max - A[i];  
        }
        return sum;
    }
};