java - 在数组中查找重复数字时出现问题，是我的技术不好吗？

Question

我使用 HashMap 来存储每个元素的出现次数，然后迭代 HashMap 以获取重复的元素，但此解决方案感觉有些不对劲。

Firecode.io 中的问题陈述:

Write a method duplicate to find the repeated or duplicate elements in an array. This method should return a list of repeated integers in a string with the elements sorted in ascending order (as illustrated below).

duplicate({1,3,4,2,1}) --> "[1]"

duplicate({1,3,4,2,1,2,4}) --> "[1, 2, 4]"

Note: You may use toString() method to return the standard string representation of most data structures, and Arrays.sort() to sort your result.*

这是我的代码:

 public String duplicate(int[] numbers) {
      HashMap < Integer, Integer > hs = new HashMap < Integer, Integer > ();
      for (int i = 0; i < numbers.length; i++) {
           if (hs.get(numbers[i]) == null) {
                hs.put(numbers[i], 1);
           } else hs.put(numbers[i], (Integer) hs.get(numbers[i]) + 1);
      }

      int size = 0;
      for (int i: hs.keySet()) {
           if (hs.get(i) > 1) {
                size++;
           }
      }
      int j = 0;
      int[] a = new int[size];
      for (int i: hs.keySet()) {
           if (hs.get(i) > 1) {
                a[j++] = i;
           }
      }

      Arrays.sort(a);
      return Arrays.toString(a);
 }

Answer 1

这是我的做法:(出于教育目的的评论，可能不会在生产代码中包含它们。)

public String duplicate(int[] numbers) {

  // holds the items we've encountered more than once.
  // TreeSet<> keeps things in sorted order for us.
  final SortedSet<Integer> duplicates = new TreeSet<>();

  // keeps track of items we've encountered.
  final Set<Integer> encountered = new HashSet<>(); 

  // iterate over every number
  for (final int number : numbers) {
    // Add the item to encountered. Set.add() will return true if 
    // the element is new to the set.
    if (!encountered.add(number)) {
       // Since the element wasn't new, ensure this item exists in the duplicates collection.
       duplicates.add(number);
    }
  }

  return duplicates.toString();
}