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)).

Median of Two Sorted Arrays

    public double findMedianSortedArrays(int A[], int B[]) {
        //http://www.cnblogs.com/tenosdoit/p/3554479.html
        int aLen = A.length;
        int bLen = B.length;
        
        if ((aLen + bLen) % 2 != 0) { //odd
            return findKth(A, B, (aLen + bLen) / 2, 0, aLen - 1, 0, bLen - 1);
        } else {
            return (findKth(A, B, (aLen + bLen) / 2 - 1, 0, aLen - 1, 0, bLen - 1)
                    + findKth(A, B, (aLen + bLen) / 2,   0, aLen - 1, 0, bLen - 1)) * (0.5);
        }
    }
    
    public double findKth(int[] A, int[] B, int k, int aStart, int aEnd, int bStart, int bEnd) {
        
        int aLen = aEnd - aStart + 1;
        int bLen = bEnd - bStart + 1;
        
        if (aLen == 0)
            return B[bStart + k];
        
        if (bLen == 0)
            return A[aStart + k];
            
        if (k == 0) 
            return A[aStart] < B[bStart] ? A[aStart] : B[bStart];
        
        int aMid = aLen * k / (aLen + bLen);    //a's middle count
        int bMid = k - aMid - 1;    //b's middle count
        
        aMid = aStart + aMid;
        bMid = bStart + bMid;
        
        if (A[aMid] > B[bMid]) {
            k = k - (bMid - bStart + 1);
		    aEnd = aMid;
		    bStart = bMid + 1;
        } else {
            k = k - (aMid - aStart + 1);
            bEnd = bMid;
            aStart = aMid + 1;
        }
        
        return findKth(A, B, k, aStart, aEnd, bStart, bEnd);
    }