剑指offer-数组中重复的数字

 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                 int s[]=new int[length];
12         for(int i=0;i<length;i++){
13             s[numbers[i]]++;
14             if(s[numbers[i]]>1){
15                 duplication[0]=numbers[i];
16                 return true;
17             }
18         }
19         return false;
20     }
21 }