找出两个含有相同元素个数的递增数列中第n小的数

最新推荐文章于 2022-01-13 00:14:33 发布

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本文介绍了一种O(log n)时间复杂度的算法，用于从两个已排序的数组中找到第n小的元素及中位数。通过递归二分查找的方式，在两个数组间高效定位目标元素。

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1.Description

You are given two arrays L1 and L2, each with n keys sorted in increasing order. For simplicity,you may assume that the keys are all distinct from each other.

(a) Describe an

O(logn) time algorithm to find the

nth smallest of the

2n keys.

(b) Describe an

O(logn) time algorithm to find the

n2th smallest (i.e., the median) of the

2n keys assuming that

n is even.

2.Java implementation

public class BinarySearch
{
	public static int Find(int seq1[],int seq2[],int left1,int right1,int left2,int right2,int key)
	{
		int mid1=(left1+right1)/2;
		int mid2=(left2+right2)/2;
		
		if(left1>right1)
		{
			int index=left2+key-1;
			return seq2[index];
		}
		if(left2>right2)
		{
			int index=left1+key-1;
			return seq1[index];
		}
		if(key<=(mid1-left1+mid2-left2+1))
		{
			if(seq1[mid1]<=seq2[mid2])
			{
				right2=mid2-1;
				return Find(seq1,seq2,left1,right1,left2,right2,key);
			}
			else
			{
				right1=mid1-1;
				return Find(seq1,seq2,left1,right1,left2,right2,key);
			}
		}
		else 
		{
			//remove the elements which is smaller than the median value
			if(seq1[mid1]<=seq2[mid2])
			{
				int keyChange=mid1+1-left1;
				left1=mid1+1;
				key=key-keyChange;
				return Find(seq1,seq2,left1,right1,left2,right2,key);
			}
			else
			{
				int keyChange=mid2+1-left2;
				left2=mid2+1;
				key=key-keyChange;
				return Find(seq1,seq2,left1,right1,left2,right2,key);
			}
		}
	}
	public static void main(String[] args)
	{
		int seq1[]={2,3,5,8,10};
		int seq2[]={4,6,7,9,12};
		int left1=0,left2=0,right1=seq1.length-1,right2=seq2.length-1;
		int x1=seq1.length;
		int x2=seq1.length/2;
		int x1Result=Find(seq1,seq2,left1,right1,left2,right2,x1);
		System.out.println("the "+x1+"th smallest number is "+x1Result);
		int x2Result=Find(seq1,seq2,left1,right1,left2,right2,x2);
		System.out.println("the "+x2+"th smallest number is "+x2Result);
	}
	
}

Screenshot of the result: