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!

Analyse: The water can be trapped equals to the min(left[i], right[i]) - height[i], whereas left[i] and right[i] is the largest number of the left and right part of the number. 

Runtime: 8ms.

class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();
        if(n <= 1) return 0;
        
        int *left = new int [n];
        int max = height[0];
        for(int i = 1; i < n - 1; i++){//find the largest of the left of a element
            left[i] = max;
            if(height[i] > max) 
                max = height[i];
        }
        max = height[n - 1];
        int result = 0;
        for(int j = n - 2; j >= 0; j--){
            int diff = min(max, left[j]) - height[j]; //the water trapped at index i
            if(diff > 0) 
                result += diff;
            if(height[j] > max) 
                max = height[j];
        }
        return result;
    }
};