Trapping Rain Water

Question:

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:

class Solution {
public:
    int trap(vector<int>& height) {
    int n=height.size();
    int sum=0;
    int *max_left=new int[n];
    int *max_right=new int[n];
    max_left[0]=0;
    max_right[n-1]=0;
    for(int i=1;i<n-1;i++)
    {
        max_left[i]=max(max_left[i-1],height[i-1]);
        max_right[n-1-i]=max(max_right[n-i],height[n-i]);
    }
    for(int j=1;j<n-1;j++)
    {        
        int val=min(max_left[j],max_right[j]);
        if(val>height[j])
            sum+=val-height[j];
    }
    return sum;   
    }
};