1 public class Solution {
 2     // Parameters:
 3     //    numbers:     an array of integers
 4     //    length:      the length of array numbers
 5     //    duplication: (Output) the duplicated number in the array number,length of duplication array is 1,so using duplication[0] = ? in implementation;
 6     //                  Here duplication like pointor in C/C++, duplication[0] equal *duplication in C/C++
 7     //    这里要特别注意~返回任意重复的一个，赋值duplication[0]
 8     // Return value:       true if the input is valid, and there are some duplications in the array number
 9     //                     otherwise false
10     public boolean duplicate(int numbers[],int length,int [] duplication) {
11        boolean[] k = new boolean[length];
12             for(int i = 0; i < k.length;i++){
13                 if(k[numbers[i]] == true){
14                     duplication[0] = numbers[i];
15                     return true;
16                 }
17                 k[numbers[i]] = true;
18             }
19            return false;
20     }
21 }