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• 原题链接：82. 删除排序链表中的重复元素 II

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1- 思路

• 1. 定义虚拟头结点
• 2. 定义 cur 指针，cur指针始终指向虚拟头结点，依此操作 cur.next 和 cur.next.next
• 3. while条件为依此判断两个连续的结点，while(cur.next!=null && cur.next.nex!=null)
• 4. 如果当前后两个结点出现重复，则一直去重到没有重复元素为止
• 5. 如果当前后两个结点没有重复元素，则直接移动 cur

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2- 题解

⭐删除排序链表中的重复元素 II——题解思路

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head==null){
            return head;
        }
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
        ListNode cur = dummyHead;
        while(cur.next!=null && cur.next.next!=null){
            // 1. 重复
            if(cur.next.val == cur.next.next.val){
                // 一直去重
                int x = cur.next.val;
                while(cur.next!=null && x==cur.next.val){
                    cur.next = cur.next.next;
                }
            }else{
                cur = cur.next;
            }
        }
        return dummyHead.next;
    }
}

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3- ACM模式

import java.util.List;
import java.util.Scanner;

public class deleteDuplicates {

    public static class ListNode{
        int val;
        ListNode next;
        ListNode(){}
        ListNode(int x){
            val = x;
        }
    }

    public static ListNode deletDuplicates(ListNode head){
        if(head==null){
            return head;
        }
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
        ListNode cur = dummyHead;
        while(cur.next!=null && cur.next.next!=null){
            if(cur.next.val==cur.next.next.val){
                int x = cur.next.val;
                while(cur.next!=null && x==cur.next.val){
                    cur.next = cur.next.next;
                }
            }else{
                cur = cur.next;
            }
        }
        return dummyHead.next;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("输入链表长度");
        int n = sc.nextInt();
        ListNode head=null,tail=null;
        for(int i = 0 ; i < n;i++){
            ListNode newNode = new ListNode(sc.nextInt());
            if(head==null){
                head = newNode;
                tail = newNode;
            }else{
                tail.next = newNode;
                tail = newNode;
            }
        }
        ListNode forRes = deletDuplicates(head);
        System.out.println("删除后的结果为");
        while (forRes!=null){
            System.out.print(forRes.val+" ");
            forRes = forRes.next;
        }
    }
}