[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

?





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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]

时间： 2024-12-25 22:20:18

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