给定一个已排序的链表的头?head
?,?删除原始链表中所有重复数字的节点,只留下不同的数字?。返回?已排序的链表?。
示例 1:
输入:head = [1,2,3,3,4,4,5] 输出:[1,2,5]
示例 2:
输入:head = [1,1,1,2,3] 输出:[2,3]
提示:
[0, 300]
?内-100 <= Node.val <= 100
/**
* 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) {
ListNode dum=new ListNode(0,head);
ListNode cur=dum;
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 dum.next;
}
}