[剑指offer] 数组中重复的数字

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++
    //    这里要特别注意~返回任意重复的一个，赋值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(i != numbers[i]){
                if(numbers[i] == numbers[numbers[i]]){
                    duplication[0] = numbers[i];
                    return true;
                }else{
                    int temp = numbers[numbers[i]];
                    numbers[numbers[i]] = numbers[i];
                    numbers[i] = temp;
                }
            }
        }
        return false;
    }
}