问题描述：

给定一个排序链表，删除所有重复的元素，使得每个元素只出现一次。

示例 1:

输入: 1->1->2输出: 1->2

示例 2:

输入: 1->1->2->3->3输出: 1->2->3

 

方法1：（超时）

1 class Solution(object): 2     def deleteDuplicates(self, head): 3         """ 4         :type head: ListNode 5         :rtype: ListNode 6         """ 7          8         p = head 9         if p == None or p.next == None:10             return head11         while p.next != None:12             q = p.next13             if p.val == q.val:14                 q = q.next15             else:16                 p.next = q17                 p = q18         return head

方法2:

1 class Solution(object): 2     def deleteDuplicates(self, head): 3         """ 4         :type head: ListNode 5         :rtype: ListNode 6         """ 7          8         p = head 9         if p == None or p.next == None:10             return head11         while p.next != None:12             q = p.next13             if p.val == q.val:14                 p.next = q.next15             else:16                 p = p.next17         return head

同上：

1 class Solution(object): 2     def deleteDuplicates(self, head): 3         """ 4         :type head: ListNode 5         :rtype: ListNode 6         """ 7         #此为不带头结点的链表 8         if head is None:#链表为空 9             return head10         cur=head11         while cur.next:#下一节点不为空12             if cur.val==cur.next.val:#第一次判断，头元素与头元素下一节点的值是否相等。。。13                 cur.next=cur.next.next14             else:15                 cur=cur.next16         return head

方法2：

1 class Solution(object): 2     def deleteDuplicates(self, head): 3         """ 4         :type head: ListNode 5         :rtype: ListNode 6         """ 7         a=[] 8         l=head 9         while l:10             if l.val in a:11                 p.next=l.next12             else:13                 a.append(l.val)14                 p=l15             l=l.next16         return head

2018-07-25 13:08:38