英文:
How to remove duplicates and matching original elements from list using java
问题
我已经尝试过使用Set集合来移除重复项，但这并不能解决我在这里提出的问题。
我正在尝试完成以下任务：我有一个列表
Integer [] list1 = {1,2,3,3};
我想要移除重复项，但我还想移除匹配的成对项，所以我想要的结果是
{1,2}
英文:
I have already tried using a Set Collection to remove duplicates but this does not solve what i am asking here.
I am trying to accomplish the following task: I have a list
Integer [] list1 = {1,2,3,3};
i want to remove the duplicates but also I want to remove the matching pair, so the result i want is
{1,2}
答案1
得分: 3
我会使用频率映射来筛选出频率大于2的元素。
- 频率映射
freqMap的条目为：{1=1, 2=1, 3=2}，我们只需要筛选出值大于2的条目.filter(e -> e.getValue() < 2)，在这里是{3=2}。 - 现在我们有了条目：
{1=1, 2=1}，但我们只想要键{1, 2}，所以我们使用map()函数来获取键：.map(e -> e.getKey())
尝试这样做：
Map<Integer, Long> freqMap = Arrays.stream(list1).collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));int[] result = freqMap.entrySet().stream().filter(e -> e.getValue() < 2).map(e -> e.getKey()).mapToInt(m -> m).toArray();System.out.println(Arrays.toString(result));
输出：
[1, 2]
英文:
I would use a frequency map to filter the elements having a frequency greater than 2.
- The frequency map
freqMaphas the entries:{1=1, 2=1, 3=2}, we just need to filter the entries having a value greater than 2.filter(e -> e.getValue()<2)which is{3=2}here. - Now we have the entries:
{1=1, 2=1}But we want only the keys{1, 2}so we usemap()function to get the keys:.map(e -> e.getKey())
Try this:
Map<Integer, Long> freqMap = Arrays.stream(list1).collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));int[] result = freqMap.entrySet().stream().filter(e -> e.getValue() < 2).map(e -> e.getKey()).mapToInt(m -> m).toArray();System.out.println(Arrays.toString(result));
Output:
[1, 2]
答案2
得分: 1
你可以使用一个Map。计算元素在一个<int, int>的映射中出现的次数，然后只获取那些计数为1的键。
所以，这就是你的示例内容，
{
1: 1,
2: 1,
3: 2,
}
然后只提取键1和2，因为它们的计数为1。时间和空间复杂度为O(n)。
英文:
You can use a Map. Count the number of times the elements appear in a <int, int> map and get only those keys that have a count of 1.
so then this is what you have for your example,
{1: 1,2: 1,3: 2,}
Then just extract keys 1 and 2 since they have a count of 1. O(n) time and space.
答案3
得分: 0
我的第一个想法是从列表中获取一个集合，然后移除列表中所有在集合中的元素，接着移除集合中所有在剩余列表中的元素。
Integer[] arr = { 1, 2, 3, 3 };List<Integer> list = new ArrayList<>(Arrays.asList(arr));Set<Integer> set = new HashSet<>(list);for (Iterator<Integer> iterator = set.iterator(); iterator.hasNext();) {list.remove(iterator.next());}set.removeAll(list);System.out.println(list);System.out.println(set);
输出：
[3][1, 2]这个方法适用于 Java 8 之前的版本。否则，只需使用 Hülya 给出的 freqMap() 方法。<details><summary>英文:</summary>My first thought was to get a set from the list, then remove all elements in the list that are in the set, and remove all elements in the set that are in the remaining list.Integer[] arr = { 1, 2, 3, 3 };List<Integer> list = new ArrayList<>(Arrays.asList(arr));Set<Integer> set = new HashSet<>(list);for (Iterator<Integer> iterator = set.iterator(); iterator.hasNext();) {list.remove(iterator.next());}set.removeAll(list);System.out.println(list);System.out.println(set);Output:[3][1, 2]This will work with versions of Java less than Java 8. Otherwise just use freqMap() as answered by Hülya.</details>
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