[LeetCode] Median of Two Sorted Arrays

Problems:
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)).
Solution :O(log(m+n))

// This is reference from Internet and I think it is good enough

public class Solution {
  public double findMedianSortedArrays(int A[], int B[]) {
    int m = A.length;
    int n = B.length;
    
    if ((m + n) % 2 != 0)
      return  findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1);
    else {
      return (findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1) + 
              findKth(A, B, (m + n) / 2 - 1, 0, m - 1, 0, n - 1))/2;
    }
  }

  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;

    // Handle special cases
    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

    // make aMid and bMid to be array index
    aMid = aMid + aStart;
    bMid = bMid + bStart;

    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);
  }
}