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))..

LeetCode-Median of Two Sorted Arrays

C++:
class Solution {
public:
    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;
    }
    double findKth(int A[],int m,int B[],int n,int k){
        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);
        int 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];
    }
};
Java:大数据通不过
public class Solution {
    public double findMedianSortedArrays(int A[], int B[]) {
	// Start typing your Java solution below
	// DO NOT write main() function
	int k=A.length+B.length;
	if(k%2==1) return findKth(A,A.length,B,B.length,k/2+1);
	else return (findKth(A,A.length,B,B.length,k/2)+findKth(A,A.length,B,B.length,k/2+1))/2;
	}
    public double findKth(int A[],int m,int B[],int n,int k){        
         if(m>n) return findKth(B,n,A,m,k);     
         if(m==0) return B[k-1];
	if(k<=1) return Math.min(A[0],B[0]);
	int pa=Math.min(k/2,m);
	int pb=k-pa;
         if(A[pa-1]<B[pb-1]) return findKth(Arrays.copyOfRange(A,pa,m),m-pa,B,n,k-pa);
	else if(A[pa-1]>B[pb-1]) return findKth(A,m,Arrays.copyOfRange(B,pb,n),n-pb,k-pb);
	else return A[pa-1];
    }
}