Leetcode: Trapping Rain Water in C++

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!

Solution:
int trap(int A[], int n) {
       vector<int> left(n, 0);
       int max = 0;
       for(int i=0;i<n;i++)
       {
           if(A[i]>max)
           {
               max = A[i];
           }
           left[i] = max;
       }
       max = 0;
       int right = 0;
       int result = 0;
       for(int i=n-1;i>=0;i--)
       {
           if(A[i]>max)
           {
               max = A[i];
           }
           right = left[i]>max?max:left[i];
           result +=right - A[i];
       }
       return result;
    }