目录

1.删除链表中所有值为val的节点 

2.反转单链表 

3.返回链表的中间节点 

4.返回链表倒数第k个节点 

5.按次序合并链表

6.按值分割链表 

7.判断链表是否为回文 

1.删除链表中所有值为val的节点 

1. 删除链表中所有值为val的节点https://leetcode.cn/problems/remove-linked-list-elements/description/

解题思路：

 解题代码：

/*** 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 cur = head.next;//从cur开始，进行节点的比较while(cur != null) {//比较节点与给定值if(cur.val == val) {//相等prev.next = cur.next;   //从cur的下一个节点开始继续与val的值进行比较，如果相同，继续向后移动curcur = cur.next;} else {//不相等prev = cur;cur = cur.next;}}if(head.val == val) {head = head.next;}return head;}
}

2.反转单链表 

2.反转单链表https://leetcode.cn/problems/reverse-linked-list/description/

j解题思路：

 解题代码：

/*** 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) {//原地反转if(head == null || head.next == null) {return head;}ListNode cur = head.next;head.next = null;   //反转后现在头结点指向节点的next域应为空while(cur != null) {ListNode curNext = cur.next;//将cur指向的节点的next域指向当前的头结点cur.next = head;//将头结点的指向修改为cur节点head = cur;//将cur指向修改为curNext的指向cur = curNext;//重复上述操作}//反转结束return head;}
}

3.返回链表的中间节点 

3. 返回链表的中间节点https://leetcode.cn/problems/middle-of-the-linked-list/description/

 解题思路：

 解题代码：

/*** 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 || head.next == null) {return head;}//定义快慢引用，初始化为引用head节点ListNode fast = head;ListNode slow = head;while(fast != null && fast.next != null) {//fast向后移动两次fast = fast.next.next;//slow向后移动一次slow = slow.next;}return slow;}
}

4.返回链表倒数第k个节点 

4. 返回链表倒数第K个节点https://www.nowcoder.com/practice/529d3ae5a407492994ad2a246518148a?tpId=13&&tqId=11167&rp=2&ru=/activity/oj&qru=/ta/coding-interviews/question-ranking

解题思路：

解题代码：

    /*public class ListNode {int val;ListNode next = null;ListNode(int val) {this.val = val;}}*/public class Solution {public ListNode FindKthToTail(ListNode head,int k) {if(head == null || k <= 0) {return null;}ListNode first = head;ListNode second = head;for(int i=0;i<k-1;i++) {//first引用向后移动k-1次// 如果在这个过程中first引用的值变为空，说明k的值大于了链表的长度，直接返回nullfirst = first.next;if(first == null) {return null;} }//两个指针一起移动，当first指针的next域变为空时，second引用的节点就是倒数第k个节点while(first.next != null) {first = first.next;second = second.next;}return second;}}

5.按次序合并链表

5. 按照升序合并两个升序链表（不能创建新的节点）https://leetcode.cn/problems/merge-two-sorted-lists/description/

解题思路：

 解题代码

/*** 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 mergeTwoLists(ListNode list1, ListNode list2) {if(list1 == null) {return list2;}if(list2 == null) {return list1;}//定义一个新的头节点。组装链表ListNode newHead = new ListNode();ListNode cur = newHead;while(list1 != null && list2 != null) {if(list1.val > list2.val) {cur.next = list2;list2 = list2.next;} else {cur.next = list1;list1 = list1.next;                }cur = cur.next;}if(list1 != null) {cur.next = list1;} else if(list2 != null) {cur.next = list2;}return newHead.next;}
}

6.按值分割链表 

6. 编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大于或等于x的结点之前https://www.nowcoder.com/practice/0e27e0b064de4eacac178676ef9c9d70?tpId=8&&tqId=11004&rp=2&ru=/activity/oj&qru=/ta/cracking-the-coding-interview/question-ranking

解题思路：

解题代码：

import java.util.*;/*
public class ListNode {int val;ListNode next = null;ListNode(int val) {this.val = val;}
}*/
public class Partition {public ListNode partition(ListNode pHead, int x) {if (pHead == null || pHead.next == null) {return pHead;}ListNode smallHead = new ListNode(-1);ListNode bigHead = new ListNode(-1);ListNode curSamll = smallHead;ListNode curBig = bigHead;//进行值的比较，节点的连接while(pHead != null) {if(pHead.val < x) {curSamll.next = pHead;curSamll = pHead;} else {curBig.next = pHead;curBig = pHead;}pHead = pHead.next;}//连接结束后，应当将bigHead这条链表的尾节点的next域置为空。防止出现死循环curBig.next = null;//连接smallHead和bigHead这两条链表curSamll.next = bigHead.next;return smallHead.next;}
}

7.判断链表是否为回文 

7.判断链表是否为回文链表http:// https://www.nowcoder.com/practice/d281619e4b3e4a60a2cc66ea32855bfa?tpId=49&&tqId=29370&rp=1&ru=/activity/oj&qru=/ta/2016test/question-ranking

解题思路：

解题代码：

import java.util.*;/*
public class ListNode {int val;ListNode next = null;ListNode(int val) {2this.val = val;}
}*/
public class PalindromeList {public boolean chkPalindrome(ListNode head) {if (head == null || head.next == null) {return true;}//寻找链表的中间节点ListNode fast = head;ListNode slow = head;while (fast != null && fast.next != null) {fast = fast.next.next;slow = slow.next;}//slow即为中间节点//从中间节点开始反转链表ListNode cur = slow.next;while(cur != null) {ListNode curNext = cur.next;cur.next = slow;slow = cur;cur = curNext;}//反转链表结束//从首个节点开始与从反转后的中间节点开始进行比较，判断是否为回文while(head.next != slow.next) {if(head.next == slow) {return true;}if(head.val != slow.val) {return false;}head = head.next;slow = slow.next;}return true;}
}