剑指offer--数组中重复的数字

题目描述

在一个长度为n的数组里的所有数字都在0到n-1的范围内。 数组中某些数字是重复的，但不知道有几个数字是重复的。也不知道每个数字重复几次。请找出数组中任意一个重复的数字。 例如，如果输入长度为7的数组{2,3,1,0,2,5,3}，那么对应的输出是第一个重复的数字2。

public class 数组中重复的数字开始 {public static void main(String[] args) {// TODO Auto-generated method stub//int[] numbers = {2,3,1,0,2,5,3};int[] numbers = {};int length =numbers.length;int[] duplication = new int[1];System.out.println(duplicate(numbers, length, duplication)+":"+duplication[0]);}    public static boolean duplicate(int numbers[],int length,int [] duplication) {    if (length!=numbers.length||numbers.length<2||numbers==null) {    duplication[0] = -1;return false;}    int[] copynum = new int[length];    for (int i = 0; i < copynum.length; i++) {    copynum[i] = -1;    if (numbers[i]>length-1) {return false;}}    for (int i = 0; i < numbers.length; i++) {if (copynum[numbers[i]]==-1) {copynum[numbers[i]]=1;}else {copynum[numbers[i]]++;}}    for (int i = 0; i < copynum.length; i++) {if (copynum[i]>1) {duplication[0]=i;return true;}}return false;    }}

别人的做法：

    public boolean duplicate(int numbers[], int length, int[] duplication) {        boolean[] k = new boolean[length];        for (int i = 0; i < k.length; i++) {            if (k[numbers[i]] == true) {                duplication[0] = numbers[i];                return true;            }            k[numbers[i]] = true;        }        return false;    }