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

时间：2014-05-23 03:59:43 收藏：0 阅读：364

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.

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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;
         
         
    }
};

[LeetCode] [Trapping Rain Water 2012-03-10],布布扣,bubuko.com