算法1

时间复杂度$O(2n)$

先走一遍，查找所有重复的元素，用一个集合set来存储
再从头走一遍，若节点的值不在集合set中，将其添加到新的链表中

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def deleteDuplication(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        mask = set()
        dummy = ListNode(0)
        dummy.next = head

        while head: # 先走一遍，将重复元素加入到集合mask中
            if head.next and head.next.val == head.val:
                mask.add(head.val)
                head = head.next.next
            else:
                head = head.next

        head = dummy.next
        res = dummy
        res.next = None
        while head: # 再走一遍，将不在集合mask中的元素加入到链表中
            if head.val in mask: # 若该节点是重复节点，不加
                head = head.next
            else:
                current = head.next
                head.next = res.next
                res.next = head
                head = current
                res = res.next
        return dummy.next

算法2

双指针 $O(n)$

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def deleteDuplication(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """

        dummy = ListNode(0)
        dummy.next = head

        p = dummy
        while p.next:
            q = p.next
            while q and p.next.val == q.val:
                q = q.next
            if p.next.next == q:
                p = p.next
            else:
                p.next = q
        return dummy.next