大家好呀，本博客目的在于记录暑假学习打卡，后续会整理成一个专栏，主要打算在暑假学习完数据结构，因此会发一些相关的数据结构实现的博客和一些刷的题，个人学习使用，也希望大家多多支持，有不足之处也请指出，谢谢大家。

一，203.移除链表元素

. - 力扣（LeetCode）

方法一：定义新链表

/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode() {}* ListNode(int val) { this.val = val; }* ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode removeElements(ListNode head, int val) {if (head == null)return head;ListNode newtail = head;ListNode newhead = new ListNode();ListNode cur = newhead;while (newtail != null) {if (newtail.val != val) {cur.next = newtail;cur = newtail;newtail = newtail.next;} else {if (newtail.next == null)cur.next = null;newtail = newtail.next;}}return newhead.next;}
}

方法二：在原链表基础上删除

/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode() {}* ListNode(int val) { this.val = val; }* ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode removeElements(ListNode head, int val) {if (head == null)return head;ListNode prev = head;ListNode pcur = head.next;while (pcur != null) {if (pcur.val == val) {prev.next = pcur.next;pcur = pcur.next;} else {prev = pcur;pcur = pcur.next;}}if(head.val==val)head=head.next;return head;}
}

二，206.反转链表

. - 力扣（LeetCode）

采用了把后面的节点头插到头节点前面的方式

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

三，876链表的中间节点

. - 力扣（LeetCode）

运用了快慢指针

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

四，面试题02.02 返回倒数第k个节点

. - 力扣（LeetCode）

采用快慢指针

/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode(int x) { val = x; }* }*/
class Solution {public int kthToLast(ListNode head, int k) {if(head==null)return -1;ListNode fast=head;ListNode slow=head;int count=0;while(count!=k-1){fast=fast.next;count++;}while(fast.next!=null){fast=fast.next;slow=slow.next;}return slow.val;}
}

本期博客就到这里，感谢阅读