[LeetCode] [Trapping Rain Water 2012-03-10]

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!

 

the key problem is Get B[i] and C[i] ,where B[i] means the max forward, and C[i] means max backward.

Sum += Min(B[i] ,C[i])-A[i] >0?Min(B[i] ,C[i])-A[i]:0

class Solution {
public:
    int trap(int A[], int n) {
        int* B = new int[n];
        int sum = 0;
        int max= 0;
        for(int i = 0 ;i<n;i++)
        {
            B[i]=max;
            if(max < A[i])
            {
                max=A[i];    
            }
        }
        max =0 ;
        for(int i=n-1; i>=0 ;i--)
        {
            int h = B[i];
            if(B[i] > max)
            {
                h = max;
            }
            if(A[i] < h)
            {
                sum += h-A[i];
            }
            
            if(max < A[i])
            {
                max=A[i];    
            }
        }
        delete[] B;
        return sum;
        
        
    }
};