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
;
int *max_right=new int
;
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;
}
};