（算法）两个有序数组的第k大的数

#include

using namespace std;

// if length of A && length of B >=k
// k is power of 2 
int FindKthElem_1(int A[],int aLeft,int aRight,int B[],int bLeft,int bRight,int k){
    /*
    if(aLeft>aRight)
        return B[bLeft+k-1];
    if(bLeft>bRight)
        return A[aLeft+k-1];
    */
    if(k==1){
        if(A[aLeft]>B[bLeft])
            return B[bLeft];
        else
            return A[aLeft];
    }

    int aKth=aLeft+(k>>1)-1;
    int bKth=bLeft+(k>>1)-1;

    if(A[aKth]<B[bKth])
        return FindKthElem_1(A,aKth+1,aRight,B,bLeft,bRight,(k>>1));
    else
        return FindKthElem_1(A,aLeft,aRight,B,bKth+1,bRight,(k>>1));
}

// if length of A + length of B >=k
int FindKthElem_2(int A[],int aLeft,int aRight,int B[],int bLeft,int bRight,int k){
    if(aLeft>aRight)
        return B[bLeft+k-1];
    if(bLeft>bRight)
        return A[aLeft+k-1];

    int aMid=aLeft+((aRight-aLeft)>>1);
    int bMid=bLeft+((bRight-bLeft)>>1);

    int halfLen=aMid-aLeft+bMid-bLeft+2;

    if(A[aMid]<B[bMid]){
        if(halfLen>k){
            return FindKthElem_2(A,aLeft,aRight,B,bLeft,bMid-1,k);
        }
        else{
            return FindKthElem_2(A,aMid+1,aRight,B,bLeft,bRight,k-(aMid-aLeft+1));
        }
    }
    else{
        if(halfLen>k){
            return FindKthElem_2(A,aLeft,aMid-1,B,bLeft,bRight,k);
        }
        else{
            return FindKthElem_2(A,aLeft,aRight,B,bMid+1,bRight,k-(bMid-bLeft+1));
        }
    }
}

int main(){
    int A[]={1,2,3,7,8,9};
    int B[]={4,5,6,10,11,12};

    int aLen=sizeof(A)/sizeof(A[0]);
    int bLen=sizeof(B)/sizeof(B[0]);

    int k;
    while(true){
        cout<<"Please Input k:"<<endl;
        cin>>k;
        cout<0,aLen-1,B,0,bLen-1,k) <<endl;
        cout<0,aLen-1,B,0,bLen-1,k) <<endl;
    }
    return 0;
}