【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 false
public 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 stub
int tmp = numbers[i];
numbers[i] = numbers[j];
numbers[j] = tmp;
}
}