[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
        vector<int> lefthighest(n,0);
        vector<int> righthighest(n,0);
        
        int leftH=0,rightH=0,sumWater=0;
        for(int i=0;i<n;i++)
        {
            lefthighest.at(i)=leftH;
            if(A[i] > leftH) leftH=A[i];
        }
        
        for(int i=n-1;i>=0;i--)
        {
            righthighest.at(i)=rightH;
            sumWater+=max(0, min(righthighest[i],lefthighest[i])-A[i]);
            
            if(A[i]>rightH) rightH=A[i];
        }
        return sumWater;
    }
};