數組中重複的數字

public class Solution {
    // Parameters:
    //    numbers:     an array of integers
    //    length:      the length of array numbers
    //    duplication: (Output) the duplicated number in the array number,length of duplication array is 1,so using duplication[0] = ? in implementation;
    //                  Here duplication like pointor in C/C++, duplication[0] equal *duplication in C/C++
    //    這裏要特別注意~返回任意重複的一個，賦值duplication[0]
    // Return value:       true if the input is valid, and there are some duplications in the array number
    //                     otherwise false
    public boolean duplicate(int numbers[],int length,int [] duplication) {
             if(length==0)return false;
        for(int i=0;i<length;i++){
            while(numbers[i]!=i){
                int m=numbers[i];
                if(numbers[m]==m){
                    duplication[0]=m;
                    return true;
                }
                numbers[i]=numbers[m];
                numbers[m]=m;//原先的numbers[i]
            }
        }
        return false;
    }
}