【Java-数据结构】Java 链表面试题上 “最后一公里”：解决复杂链表问题的致胜法宝

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引言：
Java链表，看似简单的链式结构，却蕴含着诸多有趣的特性与奥秘，等待我们去挖掘。它就像一个神秘的宝藏迷宫，每一个特性都是隐藏在迷宫深处的珍贵宝藏。链表的环，如同迷宫中的循环通道，一旦进入，便可能陷入无尽的循环；链表节点的唯一性与重复性，仿佛迷宫中的岔路，有的道路独一无二，有的却似曾相识；而链表的长度变化，又如同迷宫的动态扩展与收缩。在接下来的题目中，你将化身为勇敢的探险家，深入链表特性的迷宫，运用你的编程智慧，解开一个个谜题。通过检测链表的环、分析节点的重复性以及精准计算链表长度，你将逐渐揭开链表神秘的面纱，领略数据结构背后的奇妙逻辑。

1. 删除链表中等于给定值 val 的所有节点。移除链表元素

题目视图：

相关代码：

packageDemo1_22;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-01-22 * Time:20:56 */classListNode{int val;ListNode next;ListNode(int val){this.val = val;}}classRemoveLinkedListElements{publicListNoderemoveElements(ListNode head,int val){// 处理头节点为 null 的情况while(head!=null&& head.val == val){ head = head.next;}ListNode curr = head;// 遍历链表while(curr!=null&& curr.next!=null){if(curr.next.val == val){ curr.next = curr.next.next;}else{ curr = curr.next;}}return head;}publicstaticvoidmain(String[] args){// 创建链表 1 -> 2 -> 6 -> 3 -> 4 -> 5 -> 6ListNode node1 =newListNode(1);ListNode node2 =newListNode(2);ListNode node3 =newListNode(6);ListNode node4 =newListNode(3);ListNode node5 =newListNode(4);ListNode node6 =newListNode(5);ListNode node7 =newListNode(6); node1.next = node2; node2.next = node3; node3.next = node4; node4.next = node5; node5.next = node6; node6.next = node7;RemoveLinkedListElements solution =newRemoveLinkedListElements();// 调用 removeElements 方法，删除值为 6 的节点ListNode newHead = solution.removeElements(node1,6);// 打印删除节点后的链表元素ListNode curr = newHead;while(curr!=null){System.out.print(curr.val +" "); curr = curr.next;}}}

2.给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。反转链表

题目视图

题目详解代码：

packageDemo1_22;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-01-23 * Time:18:24 */classListNode{int val;ListNode next;ListNode(int val){this.val = val;}}publicclassReverseLinkedList{publicListNodereverseList(ListNode head){ListNode prev =null;ListNode current = head;while(current !=null){ListNode nextTemp = current.next; current.next = prev; prev = current; current = nextTemp;}return prev;}publicstaticvoidmain(String[] args){// 构建链表 1 -> 2 -> 3 -> 4 -> 5ListNode head =newListNode(1);ListNode node2 =newListNode(2);ListNode node3 =newListNode(3);ListNode node4 =newListNode(4);ListNode node5 =newListNode(5); head.next = node2; node2.next = node3; node3.next = node4; node4.next = node5;ReverseLinkedList solution =newReverseLinkedList();ListNode reversedHead = solution.reverseList(head);while(reversedHead !=null){System.out.print(reversedHead.val +" "); reversedHead = reversedHead.next;}}}

3.给你单链表的头结点 head ，请你找出并返回链表的中间结点。如果有两个中间结点，则返回第二个中间结点。链表的中间节点

题目视图：

题目讲解代码：

packageDemo1_22;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-01-23 * Time:18:30 */classListNode{int val;ListNode next;ListNode(int val){this.val = val;}}publicclassMiddleOfLinkedList{publicListNodemiddleNode(ListNode head){ListNode slow = head;ListNode fast = head;while(fast !=null&& fast.next !=null){ slow = slow.next; fast = fast.next.next;}return slow;}publicstaticvoidmain(String[] args){// 构建链表 1 -> 2 -> 3 -> 4 -> 5ListNode head =newListNode(1);ListNode node2 =newListNode(2);ListNode node3 =newListNode(3);ListNode node4 =newListNode(4);ListNode node5 =newListNode(5); head.next = node2; node2.next = node3; node3.next = node4; node4.next = node5;MiddleOfLinkedList solution =newMiddleOfLinkedList();ListNode middle = solution.middleNode(head);System.out.println(middle.val);}}

4.实现一种算法，找出单向链表中倒数第 k 个节点。返回该节点的值。返回倒数第k个结点的值

题目视图：

题目详解代码：

packageDemo1_22;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-01-23 * Time:18:33 */classListNode{int val;ListNode next;ListNode(int val){this.val = val;}}publicclassFindKthFromEnd{publicintfindKthFromEnd(ListNode head,int k){ListNode slow = head;ListNode fast = head;// 先让fast指针前进k步for(int i =0; i < k; i++){if(fast ==null){return-1;// 链表长度小于k，可根据实际情况调整返回值} fast = fast.next;}// 然后slow和fast同时前进，直到fast到达链表末尾while(fast !=null){ slow = slow.next; fast = fast.next;}return slow.val;}publicstaticvoidmain(String[] args){// 构建链表 1 -> 2 -> 3 -> 4 -> 5ListNode head =newListNode(1);ListNode node2 =newListNode(2);ListNode node3 =newListNode(3);ListNode node4 =newListNode(4);ListNode node5 =newListNode(5); head.next = node2; node2.next = node3; node3.next = node4; node4.next = node5;FindKthFromEnd solution =newFindKthFromEnd();int k =2;int result = solution.findKthFromEnd(head, k);System.out.println("链表中倒数第 "+ k +" 个节点的值是: "+ result);}}

5.将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

合并两个升序链表

题目视图：

题目详解代码：

packageDemo1_22;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-01-23 * Time:18:36 */classListNode{int val;ListNode next;ListNode(int val){this.val = val;}}publicclassMergeTwoSortedLists{publicListNodemergeTwoLists(ListNode list1,ListNode list2){ListNode dummy =newListNode(0);ListNode current = dummy;while(list1 !=null&& list2 !=null){if(list1.val < list2.val){ current.next = list1; list1 = list1.next;}else{ current.next = list2; list2 = list2.next;} current = current.next;}if(list1 !=null){ current.next = list1;}if(list2 !=null){ current.next = list2;}return dummy.next;}publicstaticvoidmain(String[] args){// 构建链表1: 1 -> 2 -> 4ListNode list1 =newListNode(1); list1.next =newListNode(2); list1.next.next =newListNode(4);// 构建链表2: 1 -> 3 -> 4ListNode list2 =newListNode(1); list2.next =newListNode(3); list2.next.next =newListNode(4);MergeTwoSortedLists solution =newMergeTwoSortedLists();ListNode mergedList = solution.mergeTwoLists(list1, list2);while(mergedList !=null){System.out.print(mergedList.val +" "); mergedList = mergedList.next;}}}

今天的链表题目就到这了，还有五到下篇见；