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 == )
            return B[k-];
        if (k <= )
            return min(A[], B[]);

        int pa = min(k / , m), pb = k - pa;
        if (A[pa-] < B[pb-])
        {
            return findKth(A + pa, m - pa, B, n, k - pa);
        }
        else if(A[pa-] > B[pb-])
        {
            return findKth(A, m, B + pb, n - pb, k - pb);
        } else
            return A[pa-];
    }

    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 /  + );
        } else
        {
            return (findKth(A, m, B, n, k / ) + findKth(A, m, B, n, k /  + )) / ;
        }
    }
};