英文:
Given n linked list containing integers, find the common elements between them and return the list
问题
我在最近的面试中被问到了这个问题。在规定的时间内，我无法想出一个令人满意的答案。
给定 N 个包含整数的单链表，找出它们之间的公共元素或交点，并返回仅包含公共元素的链表。
示例：
输入：
list1 = [1, 3, 4, 5, 10, 3]
list2 = [1, 2, 4, 8, 7, 2]
list3 = [9, 11, 1, 4]
输出：
list = [1, 4]
注意： 请注意，同一链表中的重复元素不应予考虑。
英文:
I was asked this question in one of my recent interviews. I could not come up with a satisfactory answer within the given time.
Given N singly-linked lists containing integers, find the common elements or intersection points between them and return the list containing only the common elements.
Example:
Input:
list1 = [1, 3, 4, 5, 10, 3]
list2 = [1, 2, 4, 8, 7, 2]
list3 = [9, 11, 1, 4]
Output:
list = [1, 4]
Note: Please note that duplicate elements within the same list shall not be considered.
答案1
得分: 1
如果列表实现了集合框架的通用List接口，可以使用retainAll方法实现：
List<Integer> list1 = Arrays.asList(1, 3, 4, 5, 10, 3);
List<Integer> list2 = Arrays.asList(1, 2, 4, 8, 7, 2);
List<Integer> list3 = Arrays.asList(9, 11, 1, 4);
List<Integer> result = new ArrayList<>(list1);
Stream.of(list2, list3).forEach(result::retainAll);
System.out.println(result);
输出：
[1, 4]
英文:
If the lists implement common List interface of the Collections framework, it can be implemented using retainAll method:
List<Integer> list1 = Arrays.asList(1, 3, 4, 5, 10, 3);
List<Integer> list2 = Arrays.asList(1, 2, 4, 8, 7, 2);
List<Integer> list3 = Arrays.asList(9, 11, 1, 4);
List<Integer> result = new ArrayList<>(list1);
Stream.of(list2, list3).forEach(result::retainAll);
System.out.println(result);
Output:
[1, 4]
答案2
得分: 0
步骤 1： 维护一个名为 HashMap<Integer, Integer> List_Frequency 的哈希映射，用于存储包含该元素的列表的频率。
步骤 2： 遍历每个列表。在再次遍历时，维护另一个名为 HashMap<Integer, Integer> Visit 的哈希映射，以了解您是否首次访问列表中的元素。这样，它将只导致 HashMap List_Frequency 增加一次。
步骤 3： 遍历哈希映射 List_Frequency 以了解哪些元素被访问了 N 次。
英文:
Step 1: Maintain a HashMap<Integer, Integer> List_Frequency for storing frequency of lists in which that element is present.
Step 2: Traverse each list. While traversing again maintain another HashMap<Integer, Integer> Visit to know that you are visiting an element in the list for the first time or not. So, that it leads to increment the HashMap List_Frequency only once.
Step 3: Traverse the HashMap List_Frequency to know which elements are visited N times.
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