【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 -> 6
ListNode 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 -> 5
ListNode 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 -> 5
ListNode 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 2 -> 3 -> 4 -> 5
ListNode 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 2 -> 4
ListNode list1 = new ListNode(1);
list1.next = new ListNode(2);
list1.next.next = new ListNode(4);
// 构建链表2: 1 -> 3 -> 4
ListNode 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;
}
}
}
今天的链表题目就到这了，还有五到下篇见；
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