数组中重复的数字 --剑指offer

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) {
        for(int i =0;i <length;i ++){
            int index = numbers[i] % length;
            numbers[index] += length;
            if(numbers[index] >= 2*length){
                duplication[0] = index;
                return true;
            }
        }
        return  false;
    }

}