【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：

package Demo1_22;/*** Created with IntelliJ IDEA.* Description:* User：Lenovo* Date:2025-01-22* Time:20:56*/
class ListNode {int val;ListNode next;ListNode(int val) {this.val = val;}
}class RemoveLinkedListElements {public ListNode removeElements(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;}public static void main(String[] args) {// 创建链表 1 -> 2 -> 6 -> 3 -> 4 -> 5 -> 6ListNode node1 = new ListNode(1);ListNode node2 = new ListNode(2);ListNode node3 = new ListNode(6);ListNode node4 = new ListNode(3);ListNode node5 = new ListNode(4);ListNode node6 = new ListNode(5);ListNode node7 = new ListNode(6);node1.next = node2;node2.next = node3;node3.next = node4;node4.next = node5;node5.next = node6;node6.next = node7;RemoveLinkedListElements solution = new RemoveLinkedListElements();// 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：

package Demo1_22;/*** Created with IntelliJ IDEA.* Description:* User：Lenovo* Date:2025-01-23* Time:18:24*/
class ListNode {int val;ListNode next;ListNode(int val) {this.val = val;}
}public class ReverseLinkedList {public ListNode reverseList(ListNode head) {ListNode prev = null;ListNode current = head;while (current != null) {ListNode nextTemp = current.next;current.next = prev;prev = current;current = nextTemp;}return prev;}public static void main(String[] args) {// 构建链表 1 -> 2 -> 3 -> 4 -> 5ListNode head = new ListNode(1);ListNode node2 = new ListNode(2);ListNode node3 = new ListNode(3);ListNode node4 = new ListNode(4);ListNode node5 = new ListNode(5);head.next = node2;node2.next = node3;node3.next = node4;node4.next = node5;ReverseLinkedList solution = new ReverseLinkedList();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：

package Demo1_22;/*** Created with IntelliJ IDEA.* Description:* User：Lenovo* Date:2025-01-23* Time:18:30*/
class ListNode {int val;ListNode next;ListNode(int val) {this.val = val;}
}public class MiddleOfLinkedList {public ListNode middleNode(ListNode head) {ListNode slow = head;ListNode fast = head;while (fast != null && fast.next != null) {slow = slow.next;fast = fast.next.next;}return slow;}public static void main(String[] args) {// 构建链表 1 -> 2 -> 3 -> 4 -> 5ListNode head = new ListNode(1);ListNode node2 = new ListNode(2);ListNode node3 = new ListNode(3);ListNode node4 = new ListNode(4);ListNode node5 = new ListNode(5);head.next = node2;node2.next = node3;node3.next = node4;node4.next = node5;MiddleOfLinkedList solution = new MiddleOfLinkedList();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：

package Demo1_22;/*** Created with IntelliJ IDEA.* Description:* User：Lenovo* Date:2025-01-23* Time:18:33*/
class ListNode {int val;ListNode next;ListNode(int val) {this.val = val;}
}public class FindKthFromEnd {public int findKthFromEnd(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;}public static void main(String[] args) {// 构建链表 1 -> 2 -> 3 -> 4 -> 5ListNode head = new ListNode(1);ListNode node2 = new ListNode(2);ListNode node3 = new ListNode(3);ListNode node4 = new ListNode(4);ListNode node5 = new ListNode(5);head.next = node2;node2.next = node3;node3.next = node4;node4.next = node5;FindKthFromEnd solution = new FindKthFromEnd();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：

package Demo1_22;/*** Created with IntelliJ IDEA.* Description:* User：Lenovo* Date:2025-01-23* Time:18:36*/
class ListNode {int val;ListNode next;ListNode(int val) {this.val = val;}
}public class MergeTwoSortedLists {public ListNode mergeTwoLists(ListNode list1, ListNode list2) {ListNode dummy = new ListNode(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;}public static void main(String[] args) {// 构建链表1: 1 -> 2 -> 4ListNode list1 = new ListNode(1);list1.next = new ListNode(2);list1.next.next = new ListNode(4);// 构建链表2: 1 -> 3 -> 4ListNode list2 = new ListNode(1);list2.next = new ListNode(3);list2.next.next = new ListNode(4);MergeTwoSortedLists solution = new MergeTwoSortedLists();ListNode mergedList = solution.mergeTwoLists(list1, list2);while (mergedList != null) {System.out.print(mergedList.val + " ");mergedList = mergedList.next;}}
}