141. Linked List Cycle

Given a linked list, determine if it has a cycle in it.

Follow up:

Can you solve it without using extra space?

来自 <https://leetcode.com/problems/linked-list-cycle/description/>

 

思路1:

将出现过的链表元素存在一个列表里,每次访问到下一个元素时查看列表中是否已存在

class ListNode(object):
  def __init__(self, x):
    self.val = x
    self.next = None
   
class Solution(object):
  def hasCycle(self, head):
  """
  :type head: ListNode
  :rtype: bool
  """
  # solution 1 效率很低
  # 将出现过的对象存储到列表之中
    l = []
    while head != None:
      l.append(head)
      head = head.next
      if head in l:
        return True
    return False 
 

 

思路2:

将出现过的链表元素打标签,最简单的标签就是链表的next属性,可以把每个出现过的链表元素都指向链表头,这样如果访问到指向链表头的元素,链表一定是个环(包含两种情况,一种是链表最后一个元素确实指向头元素(一次访问),另一种是链表的最后一个元素指向之前的其他非首元素(二次访问))

 1 class Solution(object):
 2     def hasCycle(self, head):
 3     """
 4     :type head: ListNode
 5     :rtype: bool
 6     """
 7         p = head
 8         while p != None:
 9             pre = p.next
10             if pre == head:
11                 return True
12             p.next = head
13             p = pre
14         return False                           

 

posted @ 2018-08-14 15:03  cwpeng  阅读(99)  评论(0编辑  收藏  举报