[Leetcode] Median of Two Sorted Arrays (Java)

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)).

一共有m+n个数，求中位数，分奇偶两种情况：

1）奇数时，中位数为第(m+n)/2+1位；

2）偶数时，中位数为第(m+n)/2位与第(m+n)/2+1位的平均数。

接下来的问题就是求第k位的数的大小：

二分法分两个部分，每个部分尽量达到k/2大小，若不够，则取最大max，与另一个数组的k-max位比较，得出结果。

public class MedianofTwoSortedArrays {
private double find(int A[],int B[],int aLow,int aHigh,int bLow,int bHigh,int k){
if(aHigh-aLow < bHigh-bLow)
return find(B, A, bLow, bHigh, aLow, aHigh, k);
if(bHigh == bLow-1)
return A[aLow+k-1];
if(k == 1)
return A[aLow]>B[bLow]?B[bLow]:A[aLow];
int bb = Math.min(k/2, bHigh-bLow+1);
int aa = k - bb;
if(B[bb+bLow-1] < A[aLow+aa-1])
return find(A, B, aLow, aHigh, bLow+bb,bHigh, k-bb);
else if(B[bb+bLow-1] > A[aLow+aa-1])
return find(A, B, aLow+aa, aHigh, bLow, bHigh, k-aa);
else
return A[aLow+aa-1];
}
public double findMedianSortedArrays(int A[], int B[]) {
int med;
boolean flag = true;
if((A.length+B.length)%2==1) {
med = (A.length+B.length)/2 +1;
}else {
med = (A.length+B.length)/2 ;
flag = false;
}
if(flag) {
if(A.length == 0 && B.length!=0)
return B[med-1];
else if (A.length!=0 && B.length == 0)
return A[med-1];
return find(A, B, 0, A.length-1, 0, B.length-1, med);
}

else {
if(A.length == 0 && B.length!=0)
return ((double)B[med]+B[med-1])/2;
else if (A.length!=0 && B.length == 0)
return ((double)A[med]+A[med-1])/2;
return (find(A, B, 0, A.length-1, 0, B.length-1, med)+find(A, B, 0, A.length-1, 0, B.length-1, med+1))/2;
}

}
public static void main(String[] args) {
int[] a = {1,3,5,7,9,11};
int[] b = {2,4,6,8,10,12,14,16};
double result = new MedianofTwoSortedArrays().findMedianSortedArrays(a, b);
System.out.println(result);
}
}