1.移除链表

class Solution {
   public ListNode removeElements(ListNode head, int val) {
       while(head !=null && head.val==val){
           head=head.next;
       }
       if(head==null){
           return head;
       }
       ListNode preNode=head;
       ListNode node=head.next;
       while(node != null){
           if(node.val==val){
               preNode.next=node.next;
           }else{
               preNode=node;
           }
           node=node.next;
       }
       return head;
    }
}

2.反转链表

class Solution {
    public ListNode reverseList(ListNode head) {
        if(head==null){
            return null;
        }
        ListNode preNode=null;
        ListNode node=head;
        while(node!=null){
            ListNode nodeNext=node.next;
            node.next=preNode;
            preNode=node;
            node=nodeNext;
        }
        return preNode;
    }
}

3.链表的中间结点

class Solution {
     public ListNode middleNode(ListNode head) {
        ListNode node=head;
        int count=0;
        while(node!=null){
            count++;
            node=node.next;
        }
        int mid=count/2+1;
        ListNode temp=head;
        for(int i=1;i<mid;i++){
            temp=temp.next;
        }
        return temp;
    }
}

4.链表中倒数第k个结点

public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
        if(head==null){
            return head;
        }
        ListNode fast=head;
        ListNode slow=head;
        for(int i=0;i<k;i++){
            if(fast==null){
                return null;
            }
            fast=fast.next;
        }
        while(fast!=null){
            slow=slow.next;
            fast=fast.next;
        }
        return slow;
    }
 }

5.合并两个有序链表

class Solution {
   public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        if(list1==null && list2==null){
            return null;
        }
        if(list1==null && list2!=null){
            return list2;
        }
        if(list1!=null && list2==null){
            return list1;
        }
        ListNode node=new ListNode(-1);
        ListNode head=node;
        ListNode temp1=list1;
        ListNode temp2=list2;
        while(temp1!=null && temp2!=null){
            if(temp1.val<temp2.val){
                head.next=temp1;
                temp1=temp1.next;
            }else{
                head.next=temp2;
                temp2=temp2.next;
            }
            head=head.next;
        }
        while(temp1!=null){
            head.next=temp1;
            temp1=temp1.next;
            head=head.next;
        }
        while(temp2!=null){
            head.next=temp2;
            temp2=temp2.next;
            head=head.next;
        }
        return node.next;
    }
}

6.链表分割

public class Partition {
       public ListNode partition(ListNode pHead, int x) {
        // write code here
        ListNode bigHead=new ListNode(-1);
        ListNode smallHead=new ListNode(-1);
        ListNode smallTail=smallHead;
        ListNode bigTail=bigHead;
        while(pHead!=null){
            if(pHead.val<x){
                smallTail.next=pHead;
                smallTail=smallTail.next;
            }else{
               bigTail.next=pHead;
                bigTail=bigTail.next;
            }
            pHead=pHead.next;
        }
        bigTail.next=null;
        smallTail.next=bigHead.next;
        return smallHead.next;
    }
}

7.删除链表中重复的节点

public class Solution {
 public static ListNode deleteDuplication(ListNode pHead) {
        if(pHead==null || pHead.next==null){
            return pHead;
        }
        ListNode root=new ListNode(-1);
        root.next=pHead;
       ListNode pre=root;
       ListNode cur=root;
       while(cur!=null){
           while(cur!=null&& cur.next!=null && cur.val==cur.next.val){
               cur=cur.next;
           }
           cur=cur.next;
           if(cur!=null && cur.next!=null && cur.val==cur.next.val){
               continue;
           }
           pre.next=cur;
           pre=pre.next;
       }
       return root.next;
    }
}

8.链表的回文结构

public class PalindromeList {
       public static boolean chkPalindrome(ListNode A) {
        // write code here
        ListNode fast=A;
        ListNode slow=A;
        //找链表的中间节点，fast一次走两步，slow一次走一步
        while(fast!=null && fast.next!=null){
            fast=fast.next.next;
            slow=slow.next;
        }
        //此时slow指的就是中间节点，现在需要进行对链表的后半部分进行链表反转
        fast=slow.next;
        while(fast!=null) {
            ListNode fastNext = fast.next;
            fast.next = slow;
            slow = fast;
            fast = fastNext;
        }
        while(slow!=A){
            if(slow.val!=A.val){
                return false;
            }
            if(A.next==slow){
                return true;
            }
            if(A.val==slow.val){
                A=A.next;
                slow=slow.next;
            }
        }
        return true;
    }
}

9.环形链表

public class Solution {
   public static  boolean hasCycle(ListNode head) {
        if(head==null || head.next==null){
            return false;
        }
        ListNode fast=head;
        ListNode slow=head;
        while(fast!=null && fast.next!=null){
            fast=fast.next.next;
            slow=slow.next;
            if(fast==slow){
                return true;
            }
        }
        return false;
    }
}

10.环形链表II

public class Solution {
     public ListNode detectCycle(ListNode head) {
        ListNode fast=head;
        ListNode slow=head;
        while(true){
            if(fast==null || fast.next==null){
                return null;
            }
            fast=fast.next.next;
            slow=slow.next;
            if(fast==slow){
                break;
            }
        }
        ListNode third=head;
        while(third!=slow){
            slow=slow.next;
            third=third.next;
        }
        return third;
    }
}