求2个有序数组都是N长的，求第K大的（假设数组没有重复）

/*
 求2有序N长数组，第K大的数
*/
#include <stdio.h>

#include <iostream>
#include <assert.h>
#include <algorithm>

#define min(x,y)  (x)<(y) ?  (x) : (y)
#define max(x,y)  (x)>(y) ?  (x) : (y)

const int N = 4;

using namespace std;


//方法一  注意 第K大，还有2个边际
int FindKth(int A[], int B[], int n, int low ,int high, int k)
{
	int m=(low+high)/2+1;
	if(low>high) return '\0';    
    if(k==m)
	{
		return A[m-1]<=B[0] ?  A[m-1] : FindKth(A, B, n, low, m-2, k);
	}
	else if(k==(N+m)) 
	{
		return A[m-1]>=B[N-1] ? A[m-1] : FindKth(A, B, n, m, high, k);
	}
    else if(k<m+1) return FindKth(A, B, n, low, high-1, k);
	else if(k>N+m) return FindKth(A, B, n, low+1, high, k);
	else if (A[m-1]>B[k-m-1]  &&  A[m-1]<B[k-m] )
	{
		return A[m-1];
	}
	else if(A[m-1]>B[k-m])
	{
		return FindKth(A,B,n,low ,m-2, k);
	}
	else  return FindKth(A, B, n, m, high, k);
}

int Find_Kth_IN_Arrays(int A[], int B[], int n, int k)
{
     int value=FindKth(A, B, n, 0, n-1, k);
	 if(value=='\0') value=FindKth(B, A, n, 0, n-1, k);
     return value;
}


//方法二，

int FindKthElement(int a[], int b[], int k)
{
    if(k == 1)
        return min(a[0], b[0]);
    else if(k == N * 2)
        return max(a[N - 1], b[N - 1]);
    int left = max(k - N - 1, 0), right = N - 1;
    while(left <= right)
    {
        int mid = (left + right) / 2;

        if(mid + 1 == k)
        {
            if(a[mid] < b[0])
                return a[mid];
            else
                right = mid - 1;
        }
        else if(mid + 1 + N == k)
        {
            if(a[mid] > b[k - mid - 2])
                return a[mid];
            else
                left = mid + 1;
        }
        else
        {
            if(a[mid] > b[k - mid - 2] && a[mid] < b[k - mid  - 1])
                return a[mid];
            else if(a[mid] > b[k - mid - 2] && a[mid] > b[k - mid - 1])
                right = mid - 1;
            else
                left = mid + 1;
        }
    }
    return FindKthElement(b, a, k);
}

int main()
{
    int K;
    int a[] = {1, 3, 7, 8};
    int b[] = {2, 4 , 5, 6};
    while(cin >> K)
    {
        assert(K >= 1 && K <= N * 2);
        cout << FindKthElement(a, b, K) << endl;
		cout<<Find_Kth_IN_Arrays(a, b, 4, K)<<endl;
    }

    return 0;
}