Linked List Cycle
题目链接
python代码实现:
# Definition for singly-linked list.
# 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
"""
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if slow == fast:
return True
return False
实现思路:
快慢指针法。定义两个指针:快指针每次走一步;慢指针每次走两步。依次循环下去,如果链表存在环,那么快慢指针一定会有相等的时候。
为了便于理解,你可以想象在操场跑步的两个人,一个快一个慢,那么他们一定会相遇(无论他们的起始点是不是在操场)