1、解法1set 判断是否visited# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val x # self.next None class Solution: def hasCycle(self, head: Optional[ListNode]) - bool: # 环形链表判断 # 节点范围[0,10e4] # 解法1哈希set visited set() if head None: return False if head.next head: return True node head while node ! None: if node in visited: return True else: visited.add(node) node node.next return False2、解法2快慢指针迟早追上# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val x # self.next None class Solution: def hasCycle(self, head: Optional[ListNode]) - bool: # 环形链表判断 # 节点范围[0,10e4] # 解法1哈希set # 解法2时间O(n),空间O(1), 快慢指针 if head None: return False if head.next head: return True # 迟早追上 slow,fast head,head while fast ! None and fast.next!None: fast fast.next.next slow slow.next if slow fast: return True return False
