题目介绍

题解

法一：带dummy node

class Solution {public ListNode deleteDuplicates(ListNode head) {//根据提示，val的值在-100~100，如果括号里面不填则默认dummy.val=0，可能会和某些测试用例的val一样导致出错ListNode dummy = new ListNode(101); dummy.next = head;ListNode cur = dummy;while(cur.next != null){if(cur.val == cur.next.val){cur.next = cur.next.next;}else{cur = cur.next;}}return dummy.next;}
}

法二：不带dummy node
本题的头结点不会被删除或修改，所以不需要dummy也可以，只需要判断一下head是否为空即可

class Solution {public ListNode deleteDuplicates(ListNode head) {if(head == null){return head;}ListNode cur = head;while(cur.next != null){if(cur.val == cur.next.val){cur.next = cur.next.next;}else{cur = cur.next;}}return head;}
}