题解 | #判断链表中是否有环#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

#
# 
# @param head ListNode类 
# @return bool布尔型
#
class Solution:
    def hasCycle(self , head: ListNode) -> bool:
        #判断一个链表里是否有环
        slow = head
        fast = head
        while slow and fast:
            slow = slow.next
            if fast.next:
                fast = fast.next.next
            else:
                return False
            if slow == fast:
                return True
        return False


        

判断一个链表中是否有环，有环使用快慢指针必定会重合。 边界条件是快指针。