Linked List Cycle

原题地址：http://oj.leetcode.com/problems/linked-list-cycle/

题意：判断链表中是否存在环路。slow指针和fast指针开始同时指向头结点head，fast每次走两步，slow每次走一步。如果链表不存在环，那么fast或者fast.next会先到None。如果链表中存在环路，则由于fast指针移动的速度是slow指针移动速度的两倍，所以在进入环路以后，两个指针迟早会相遇，如果在某一时刻slow==fast，说明链表存在环路。

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

class Solution:
# @param head, a ListNode
# @return a boolean
def hasCycle(self, head):
if head == None or head.next == None:
return False
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if slow == fast:
return True
return False

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