Question

这里是算法的伪代码，用于查找两个数组的中位数

T WO -A RRAY-M EDIAN (X, Y )
n ← length[X]                  £ n also equals length[Y ]
median ← F IND -M EDIAN (X, Y, n, 1, n)
if median = NOT- FOUND
   then median ← F IND -M EDIAN (Y, X, n, 1, n)
return median
F IND -M EDIAN ( A, B, n, low, high)
if low > high
   then return NOT- FOUND
   else k ← (low + high)/2
        if k = n and A[n] ≤ B[1]
           then return A[n]
        elseif k < n and B[n − k] ≤ A[k] ≤ B[n − k + 1]
           then return A[k]
        elseif A[k] > B[n − k + 1]
           then return F IND -M EDIAN ( A, B, n, low, k − 1)
        else return F IND -M EDIAN ( A, B, n, k + 1, high)

来自MIT算法的介绍

这是我的代码，如果我的代码中有问题告诉我

    public class findmedian{
public static  int   find(int []x,int []y){
int median;
int n=x.length;
   median=Median( x, y, n,0, n);
  if ( median<-100000){
    Median(y,x, n,1, n);

}

 return median;  


 }


public static void main(String[]args){
 int x[]=new int[]{3,5,7,8,5,9,10};
 int y[]=new int[]{12,4,8,9,6,7,10};
find(x,y);


}

public static  int  Median(int []a,int []b,int n,int low,int high)
{


   if ( low>high)
 return -1  ;
  int k=(low+high)/2;
  if (k==n  && a[n]<=b[1])
   return a[n];
 else   if ( k<n && b[n-k]<=a[k] && b[n-k]<=b[n-k+1])
 return a[k];
   else if ( a[k]> b[n-k-1])
 return   Median( a,b,n,low,k-1);





 return  Median(a,b,n,k+1,high);





}


}

编译后它不返回任何请帮助

Answer 1

 if ( low>high)
 return;

这是错误吗？您需要返回某些内容。（返回类型为int）

关于两个数组的中位数的问题

1 个答案: