题目

• 原题连接：82. 删除排序链表中的重复元素 II

1- 思路

• 模式识别1：已排序链表 ——> 判重逻辑 ，涉及到 while

2- 实现

⭐82. 删除排序链表中的重复元素 II——题解思路

class Solution {
    public ListNode deleteDuplicates(ListNode 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;
    }
}

3- ACM 实现

public class removeDuplicate {

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

    public static ListNode removeDuplicate(ListNode 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 && cur.next.val == x){
                    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;
        System.out.println("输入链表元素");
        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 res = removeDuplicate(head);
        while(res!=null){
            System.out.print(res.val+" ");
            res = res.next;
        }
    }
}