3、数组中重复的数字

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++)
{
while(numbers[i]!=i)
{
if(numbers[i]==numbers[numbers[i]])
{
duplication[0] = numbers[i];
return true;
}
swap(numbers,i,numbers[i]);
}
}
return false;
}
public void swap(int[] numbers,int num1,int num2)
{
int a=numbers[num1];
numbers[num1]=numbers[num2];
numbers[num2]=a;
}
}