解法一： HashSet

import java.util.*;
public class Solution {
    public boolean duplicate(int numbers[],int length,int [] duplication) {
        duplication[0]=-1;
        if(numbers==null||numbers.length==0) return false;
        Set<Integer> set=new HashSet<>();
        for(int i=0; i<numbers.length; i++){
            if(!set.contains(numbers[i]))
                set.add(numbers[i]);
            else{
                duplication[0]=numbers[i];
                break;
            } 
        }
        return (duplication[0]!=-1);
    }
}

解法二：in-place算法

public class Solution {
    public boolean duplicate(int numbers[],int length,int [] duplication) {
        duplication[0]=-1;
        for(int i=0; i<length; i++){
            while(numbers[i]!=i&&numbers[i]!=numbers[numbers[i]]){
                int t=numbers[i];
                int s=numbers[t];
                numbers[t]=t;
                numbers[i]=s;
            }
        }
        for(int i=0; i<length; i++){
            if(numbers[i]!=i){
                duplication[0]=numbers[i]; return true;
            }
        }
        return false;
    }
}