[leetcode] 141. Linked List Cycle @ python

最新推荐文章于 2021-04-07 20:44:25 发布

原创最新推荐文章于 2021-04-07 20:44:25 发布 · 209 阅读

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本文介绍两种检测链表中是否存在循环的有效方法。解法一：通过遍历链表并将已访问节点记录在一个列表中，检查是否有重复节点。解法二：采用快慢指针技巧，快指针每次移动两步，慢指针每次移动一步，判断两者是否相遇。

原题

https://leetcode.com/problems/linked-list-cycle/

解法1

遍历链表, 将已见过的节点放入seen, 如果遇到已见过的节点则返回True, 遍历结束如果没发现重复的节点则返回False.
Time: O(n)
Space: O(1)

代码

# 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
        """
        seen = []
        while head:
            if head in seen:
                return True
            else:
                seen.append(head)
                head = head.next
        return False

解法2

定义两个指针, fast和slow, fast每次走两步, slow每次走一步, 如果fast和slow指向同一个节点则返回True, 遍历结束则返回False.
Time: O(n/2)
Space: O(1)

代码

# 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, head
        while fast and fast.next:            
            fast = fast.next.next
            slow = slow.next
            if slow == fast:
                return True
        return False