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) {
        //way1.排序然后遍历时间O（nlogn）
        //way2.hashmap，o(n)的时间,o(n)的空间
        //way3.遍历数组和当前下标比较，并交换放到对于的下标下，直到发现重复的数。o(n)的时间,o(1)的空间
        if (numbers == null || numbers.length == 0) {
            return false;
        }
        for (int i = 0; i < numbers.length; i++) {
            //数字都在0到n-1的范围内
            if (numbers[i] < 0 || numbers[i] >= numbers.length ) {
                return false;
            }
        }
        for (int i = 0; i < numbers.length; i++) {
            //如果当前值和下标相等，就下一个
            if (numbers[i] == i) {
                continue;
            }
            //当前值和下标不等，且发现，当前值和对于下标的值相等，发现重复的数
            if (numbers[i] == numbers[numbers[i]]){
                duplication[0] = numbers[i];
                return true;
            }
            //将当前值放到对应的下标位置
            int temp = numbers[i];
            numbers[i] = numbers[temp];
            numbers[temp] = temp;
        }
        return false;
    
    }
}