Leetcode剑指Offer刷题指南：

Leetcode剑指Offer刷题-学习计划目录_DEGv587的博客-CSDN博客

剑指 Offer 18. 删除链表的节点

解法一：双指针

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) { val = x; }* }*/
class Solution {public ListNode deleteNode(ListNode head, int val) {//判空if (head == null) return head;//设定cur代表当前节点和pre代表上一个节点ListNode cur = head;ListNode pre = null;//判断首节点是否为要删除节点，如果是，直接返回下一个节点if (cur.val == val) return cur.next;//不是要删除节点就向后走while (cur.val != val) {pre = cur;cur = cur.next;}//找到了要删除节点pre.next = pre.next.next;return head;}
}

解法二：单指针

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) { val = x; }* }*/
class Solution {public ListNode deleteNode(ListNode head, int val) {if (head == null) return head;if (head.val == val) return head.next;ListNode cur = head;while (cur.next != null && cur.next.val != val) {cur = cur.next;}if (cur.next != null) {cur.next = cur.next.next;}return head;}
}

剑指 Offer 22. 链表中倒数第k个节点

解法：快慢指针

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) { val = x; }* }*/
class Solution {public ListNode getKthFromEnd(ListNode head, int k) {if (head == null) return head;ListNode slow = head;ListNode fast = head;//fast先走k步while (k > 0 && fast != null) {fast = fast.next;k--;}//slow和fast一起走while (fast != null) {slow = slow.next;fast = fast.next;}return slow;}
}