Median of Two Sorted Arrays Java

Problem: 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)).

Key to solve: Bisection + Recursion
    Convert find Median in 2 SortedArrays problem into
    find Kth number k=(m+n)/2
    There are 3 cases to locate Median during recursion process
    1. if A[k/2]==B[k/2] => A[k/2] is Median
    2. else if A[k/2]>B[k/2] => cut off elements before k/2 in B
    3. else A[k/2]<B[k/2] => cut off elements before k/2 in A
    We are able to cut off k/2 elements each time.
    when k==1 return median
    do not forget to check for special cases that including A.length==0 || B.length==0 || total==0

public class Solution {    public double findMedianSortedArrays(int A[], int B[]) {          int lenA=A.length;        int lenB=B.length;        int total=lenA+lenB;        if(total==0) return 0;        if(total%2==0){ //even length           return (helperBisection(A,B,0,lenA-1,0,lenB-1,total/2)+helperBisection(A,B,0,lenA-1,0,lenB-1,total/2+1))/2.0;        }else{  //odd length            return  helperBisection(A,B,0,lenA-1,0,lenB-1,total/2+1);        }    }    private static int helperBisection(int[] A,int[] B, int aStart, int aEnd, int bStart, int bEnd, int k){        int lenA=aEnd-aStart+1;        int lenB=bEnd-bStart+1;        //always keep A[] length less than B[]        if(lenA>lenB) return helperBisection(B,A,bStart,bEnd,aStart,aEnd,k);         //check special case only for A is good enough here since we did swap in above        if(lenA==0) return B[k+bStart-1];        //base case        if(k==1){            return Math.min(A[aStart],B[bStart]);        }        int postA=Math.min(k/2,lenA);        int postB=k-postA;        if(A[aStart+postA-1]==B[bStart+postB-1]){                return A[aStart+postA-1];        }else if(A[aStart+postA-1]>B[bStart+postB-1]){            //cut off elements before k/2 in B            return helperBisection(A,B,aStart,aStart+postA-1,bStart+postB,bEnd,k-postB);        }else {            //cut off elements before k/2 in A            return helperBisection(A,B,aStart+postA,aEnd,bStart,bStart+postB-1,k-postA);        }    }}