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) {
//鲁棒性
if(numbers == null || numbers.length != length){
duplication[0] = -1;
return false;
}
int i;
for(i = 0;i < length;i++){
if(numbers[i] < 0 || numbers[i] >= length){
duplication[0] = -1;
return false;
}
}
for(i = 0;i < length;i++){
while(numbers[i] != i){
if(numbers[i] == numbers[numbers[i]]){
duplication[0] = numbers[i];
return true;
}
int temp = numbers[i];
numbers[i] = numbers[temp];
numbers[temp] = temp;
}
}
return false;
}
}
注意: 这样写会报错，因为第三行的numbers[i]已经改变。
