leetcode 141.环形链表

生信研究猿2026-04-05 8:28

看了题解：一个指针每次跑两步，一个指针每次跑一步，如果是环形链表一定会相遇，不会错过。

python 复制代码

# 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:
        if head == None:
            return False
        fast = head.next
        slow = head
        

        while(fast != slow and fast !=None):
            if fast.next==None:
                return False
            fast = fast.next.next
            slow = slow.next
        
        return fast==slow

题解代码：

python 复制代码

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        slow = fast = head  # 乌龟和兔子同时从起点出发
        while fast and fast.next:
            slow = slow.next  # 乌龟走一步
            fast = fast.next.next  # 兔子走两步
            if fast is slow:  # 兔子追上乌龟（套圈），说明有环
                return True
        return False  # 访问到了链表末尾，无环

作者：灵茶山艾府
链接：https://leetcode.cn/problems/linked-list-cycle/solutions/1999269/mei-xiang-ming-bai-yi-ge-shi-pin-jiang-t-c4sw/
来源：力扣（LeetCode）
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