Median of Two Sorted Arrays-----LeetCode

There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

 

解题思路：

该题可以解决所有求有序数组A和B有序合并之后第k小的数！

该题的重要结论：

如果A[k/2-1]<B[k/2-1]，那么A[0]~A[k/2-1]一定在第k小的数的序列当中，可以用反证法证明。

 

具体的分析过程可以参考http://blog.csdn.net/zxzxy1988/article/details/8587244

 

class Solution {

public:

    double findKth(int A[], int m, int B[], int n, int k)

    {

        //m is equal or smaller than n

        if (m > n)

            return findKth(B, n, A, m, k);

        if (m == 0)

            return B[k-1];

        if (k <= 1)

            return min(A[0], B[0]);



        int pa = min(k / 2, m), pb = k - pa;

        if (A[pa-1] < B[pb-1])

        {

            return findKth(A + pa, m - pa, B, n, k - pa);

        }

        else if(A[pa-1] > B[pb-1])

        {

            return findKth(A, m, B + pb, n - pb, k - pb);

        } else

            return A[pa-1];

    }



    double findMedianSortedArrays(int A[], int m, int B[], int n) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        int k = m + n;

        if (k & 0x1)

        {

            return findKth(A, m, B, n, k / 2 + 1);

        } else

        {

            return (findKth(A, m, B, n, k / 2) + findKth(A, m, B, n, k / 2 + 1)) / 2;

        }

    }

};