简直痛哭流涕！！！！！
又一次做出来了链表题，而且是中等难度！！！而且效果不错！！！！
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def deleteDuplicates(self, head: ListNode) -> ListNode: if not head: return if head.next==None: return head if head.val==head.next.val and head.next.next==None: return r=head p=head.next q=p.next while p: if r.val==p.val: #print(r.val) a=r.val r=q #print(a) while r.val==a and r.next!=None: r=r.next if r.val==a: return head=r #print(r.val) if r.next!=None: p=r.next else: return head #print(p.val) if p.next!=None: q=p.next else: if r.val==p.val: return return head elif p.next!=None and p.val==q.val: b=p.val if q.next==None: r.next=None return head p=q.next while p.val==b and p.next!=None: p=p.next #print(r.val,p.val,b) if p.val==b: r.next=None return head q=p.next r.next=p #print(r.val,p.val,q.val) else: if r.next==None: return head else: r=r.next if p.next==None: return head else: p=p.next if q.next==None: return head else: q=q.next return head
执行用时为 36 ms 的范例 # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def deleteDuplicates(self, head: ListNode) -> ListNode: if not head: return head dummy = ListNode(0) dummy.next = head slow, fast = dummy, head while fast: while fast.next and fast.next.val == fast.val: fast = fast.next if slow.next == fast: slow = fast else: slow.next = fast.next fast = fast.next return dummy.next
——2019.10.23
原文：https://www.cnblogs.com/taoyuxin/p/11729322.html