数组中重复的数字【剑指】

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++){
            if(numbers[i]<0||numbers[i]>length-1){
                return false;
            }
        }

        int index=0;
        while(index<length-1){
            for(int j=index+1;j<length;j++){
                if(numbers[index]==numbers[j]){
                    duplication[0]=numbers[j];
                    return true;
                }
            }
            index++;
        }
        return false;
    }
}