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) {

boolean [] b = new boolean[length];

for(int i=0;i<b.length;i++)

{

if(b[numbers[i]]==true)

{

duplication[0]=numbers[i];

return true;

}

b[numbers[i]]=true;

}

return false;

/*

StringBuffer sb = new StringBuffer();

for(int i = 0; i < length; i++){

sb.append(numbers[i] + "");

}

for(int j = 0; j < length; j++){

//indexOf 返回的是此参数在string中的缩影

if(sb.indexOf(numbers[j]+"") != sb.lastIndexOf(numbers[j]+"")){

duplication[0] = numbers[j];

return true;

}

}

return false;

*/

}

}