Leetcode82删除排序链表中重复元素2

代码：

为了找到结果的head 可以在head前加一个dummy dummy->head 再顺序去重

/**
 * 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 deleteDuplicates(ListNode head) {
        if(head==null)return head;
        // while(head.next!=null&&head.val==head.next.val){
        //     while(n1!=null&&n1.next!=null&&n1.val==n1.next.val){
        //         if(n1.next.next!=null){
        //             head=n1.next.next;
        //         }else{
        //             return null;
        //         }
        //         n1=n1.next;
        //     }
        // }
        if(head.next==null)return head;
        ListNode dummy = new ListNode(0,head);
        ListNode n1 = dummy;
 
        while(n1.next!=null&&n1.next.next!=null){
            ListNode n2 = n1.next;
            ListNode n3 = n1.next.next;
            if(n2.val==n3.val){
                while(n2.val==n3.val){
                    n2 = n3;
                    n3 = n3.next;
                    if(n2==null||n3==null){
                        n1.next=null;
                        break;
                    }else{
                        n1.next = n3;
                    }                    
                }
            }else{
                n1=n1.next;
            }
        }
        return dummy.next;
    }
}