【剑指Offer】3. 数组中重复的数字

Input:
{2, 3, 1, 0, 2, 5}

Output:
2

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) {
        if(numbers == null || length <= 0) 
            return false;
        
        for(int i = 0; i < length; i++) {
            while(i != numbers[i]) {
                if(numbers[i] ==  numbers[numbers[i]]) {
                    duplication[0] = numbers[i];
                    return true;
                }
                swap(numbers, i, numbers[i]);
            }
        }
        return false;
    }
    
    public void swap(int arr[], int i, int j) {
        int temp = 0;
        temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }
}

import java.util.HashMap;
import java.util.Map;

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) {
        if(numbers == null || length <= 0) 
            return false;

        HashMap<Integer, Integer> map = new HashMap<>();

        for(int i = 0; i < length; i++) {
           map.put(numbers[i], map.containsKey(numbers[i]) ? map.get(numbers[i])+1 : 1);
           if (map.get(numbers[i]) == 2) {
               duplication[0] = numbers[i];
               return true;
           }
        }
        return false;
    }
    
}