使用有序Map进行判断

使用有序Map进行判断

import java.util.*;
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) {
        Map<Integer, Integer> map = new LinkedHashMap<Integer, Integer>(length);
        if(numbers==null || numbers.length==0){
            return false;
        }

        for (int i = 0; i < numbers.length; i++) {
            if (!map.containsKey(numbers[i])) {
                map.put(numbers[i], 1);
            } else {
                int value = map.get(numbers[i]);
                map.put(numbers[i], ++value);
            }
        }

        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            Integer key = entry.getKey();
            Integer value = entry.getValue();
            if (value != 1) {
                duplication[0] = key;
                return true;
            }
        }
        return false;
    }
}