Leetcode:median of Sorted Arrays

Title Requirements:

There is sorted arrays A and B of size m and N respectively. Find The median of the sorted arrays. The overall run time complexity should be O (log (m+n)).

Title Address: https://leetcode.com/problems/median-of-two-sorted-arrays/

 Public classSolution {Private Static intFind_kth (int[] A,int[] B,intk) {        //Arrays.copyofrange (original, from, to)        if(a.length>0) System.out.println ("Current A (" +a.length+ "):" +a[0]+ "--" +a[a.length-1]); if(b.length>0) System.out.println ("Current B (" +b.length+ "):" +b[0]+ "--" +b[b.length-1]); System.out.println ("Current K:" +k); if(a.length>b.length)returnfind_kth (b,a,k); if(a.length==0)returnB[k-1]; if(k==1)return(A[0]<b[0]? A[0]:b[0]); intIa= (K/2) >= (a.length)? A.length: (K/2)); intib=k-IA; System.out.println ("+a[ia-1]+" B in contrast a "+b[ib-1]); if(a[ia-1]==b[ib-1]){            returnA[ia-1]; }        Else if(a[ia-1]<b[ib-1]){            int[] Temp_a=Arrays.copyofrange (A, IA, a.length); int[] Temp_b=arrays.copyofrange (B, 0, IB); System.out.print ("Eliminate the" +a[0]+ "--" +a[ia-1]); if(ib<b.length) System.out.println ("; B "+b[ib]+"-"+b[b.length-1]); returnFind_kth (temp_a,temp_b,k-IA); }Else if(a[ia-1]>b[ib-1]){            int[] temp_b=Arrays.copyofrange (B, IB, b.length); int[] Temp_a=arrays.copyofrange (A, 0, IA); if(ia<a.length) System.out.print ("+a[ia]+" in "a" and "+a[a.length-1"]); System.out.println ("; B "+b[0]+"-"+b[ib-1]); returnFind_kth (temp_a,temp_b,k-IB); }        return0; }     Public Static DoubleFindmediansortedarrays (intA[],intb[]) {       intm=a.length,n=b.length; inttotal=m+O; if(total%2==0){           //System.out.println ("GG");           return(Find_kth (A,B,TOTAL/2) +find_kth (a,b,total/2+1))/2.0; }Else{           return(Double) find_kth (a,b,total/2+1); }            }}

The problem can be generalized to find the 2 ordered array (A, b) of the K-large elements, you can find a K/2 large element and the K/2 large element of a comparison, assuming that K/2 is an integer, there are the following 3 cases:

AK/2>BK/2: Then there must be BK/2 itself and the preceding element is not the K-large, and the elements behind the AK/2 do not include AK/2 must not be the largest, can be excluded.

But the algorithm does not need K/2 to be an integer, only uses the ia+ib=k to be able.

Leetcode:median of Sorted Arrays