编写代码,移除未排序链表中的重复节点。保留最开始出现的节点。
示例1:
输入:[1, 2, 3, 3, 2, 1] 输出:[1, 2, 3]
示例2:
输入:[1, 1, 1, 1, 2] 输出:[1, 2]
提示:
- 链表长度在[0, 20000]范围内。
- 链表元素在[0, 20000]范围内。
进阶:
如果不得使用临时缓冲区,该怎么解决?
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def removeDuplicateNodes(self, head: ListNode) -> ListNode:
if head is None or head.next is None:
return head
cache = set()
cache.add(head.val)
cur, p = head, head.next
while p:
if p.val not in cache:
cur.next = p
cur = cur.next
cache.add(p.val)
p = p.next
cur.next = None
return head
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeDuplicateNodes(ListNode head) {
if (head == null || head.next == null) {
return head;
}
Set<Integer> s = new HashSet<>();
s.add(head.val);
ListNode p = head.next, cur = head;
while (p != null) {
if (!s.contains(p.val)) {
cur.next = p;
cur = cur.next;
s.add(p.val);
}
p = p.next;
}
cur.next = null;
return head;
}
}