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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;
*/
}
}
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