Python3解leetcode Linked List Cycle

问题描述：

Given a linked list, determine if it has a cycle in it.

To represent a cycle in the given linked list, we use an integer poswhich represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

思路1：

设置两个指针，一个每次走一步，一个每次走两步，那么如果有环，两个指针一定会相遇。

但这个代码写出来仅仅beats 28.6%，效率不高

代码1:

# 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
        """
        if head == None or head.next == None: return False
        head2 = head
        while head2.next != None and head.next != None:#当两个指针的next都不为None
            if (head2.next.next != None):#如果走两步的指针head2的next.next不为None
                head2 = head2.next.next
            else:
                return False
            head = head.next
            if(head == head2):
                return True
        return False

将上述思路优化后，代码如下：

if not head  or not head.next: return False
        head2 = head
        while not head2 and not head2.next:#只需要关注走的比较快的指针是否为空即可，若较快指针不为空，则较慢指针肯定不为空
            head2 = head2.next.next
            head = head.next
            if(head == head2):
                return True
        return False  

思路2：

循环遍历整个列表，将遍历过的节点值设置为inf(最大值），如果后续遍历的节点值有等于inf的，则证明有环，否则就是没有环。

该算法写出来beats65.7%的人，还需要优化

代码2：

# 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
        """
        if head == None or head.next == None: return False
        head.val = float(‘inf‘)
        while head.next != None:#当指针的next都不为None
            if head.val == head.next.val:
                return True
            else:
                head = head.next
                head.val = float(‘inf‘)
        return False

原文地址：https://www.cnblogs.com/xiaohua92/p/11080739.html