// Create two lists List<String> list1 =newArrayList<>(); list1.add("A"); list1.add("B"); list1.add("C"); List<String> list2 =newArrayList<>(); list2.add("B"); list2.add("C"); list2.add("D"); // 寻找共同元素 List<String> commonElements =newArrayList<>(list1); com...
head2.nextNode.nextNode.nextNode = new ListNode(5); ListMeet listMeet = new ListMeet(); if(listMeet.hasIntersection(head1,head2)==true){ System.out.println(listMeet.twoLikeMeet(head1,head2).data); }else{ System.out.println("两个链表不相交"); } } 1. 2. 3. 4. 5. 6. 7. 8...
今天介绍的是LeetCode算法题中Easy级别的第37题(顺位题号是160)。编写程序以找到两个单链表交叉的节点。例如: 以下两个链表: A: a1→a2 ↘ c1→c2→c3 ↗ B:b1→b2→b3 链表A和链表B在c1处相交。 注意: 如果两个链接列表根本没有交集,则返回null。 函数返回后,链接列表必须保留其原始结构。 可以假设整个...
= list2.size()) return false; // (switch the two checks above if (null, null) should return false) Map<String, Count> counts = new HashMap<>(); // Count the items in list1 for (String item : list1) { if (!counts.containsKey(item)) counts.put(item, new C...
在main方法中,我们定义了两个字符串str1和str2,并调用hasIntersection方法来判断它们是否有交集。根据判断结果,我们输出相应的提示信息。 流程图 下面是上述代码的流程图表示: flowchart TD A(Start) --> B{Has Intersection?} B --> |Yes| C[Print "The two strings have intersection."] ...
先算出两个链表各自的长度,然后从较长的链表先遍历,遍历到较长链表剩余长度和较短链表一样时,用两个指针同时遍历两个链表。这样如果链表有交点的话,两个指针已经一定会相遇。 代码 publicclassSolution{publicListNode getIntersectionNode(ListNode headA, ListNode headB) { ...
I have a new list and an old list. In Java is there a standard way/library that allows me to compare these two lists and determine which items have been updated/deleted or are completely new? E.g. I should end up with three lists - Deleted items (items in old but not in new), ...
c1 → c2 → c3 ↗ B: b1 → b2 → b3 begin to intersect at node c1. Notes: If the two linked lists have no intersection at all, returnnull. The linked lists must retain their original structure after the function returns. You may assume there are no cycles anywhere in the entire li...
1. Comparing Two Lists for Equality 1.1. Sort then Compare The following Java program tests if two given lists are equal. To test equality, we need to sort both lists and compare both lists usingequals()method. TheList.equals()method returnstruefor two list instances if and only if: ...
5、时间和空间权衡: 此方法时间复杂度较低,但需要额外的空间存储HashSet。寻找数组交集是常见的算法问题,通过集合的方式可以有效地解决,同时考虑算法的时间和空间复杂 How to find the intersection of two arrays in Java?Use HashSet for storage: HashSet can quickly search and store unique elements.Store ...