【剑指offer】数组中重复的数字

import java.util.Arrays;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 falsepublic static boolean duplicate(int numbers[], int length, int[] duplication) {for (int i = 0; i < length; i++) {if (numbers[i] != i) {if (numbers[i] != numbers[numbers[i]]) {//处理他自己和他自己对应脚标目标数不一样的情况swap(numbers, i, numbers[i]);} else {//一样了说明它对应的数那里有人占了 有重复数字了 存入duplication[0]duplication[0] = numbers[i];return true;}}}return false;}private static void swap(int[] numbers, int i, int j) {// TODO Auto-generated method stubint tmp = numbers[i];numbers[i] = numbers[j];numbers[j] = tmp;}}