卖萌的弱渣

I am stupid, I am hungry.

Linked List Cycle

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

Follow up:

Space: O(1)

Solution

(Linked-List-Cycle.py) download
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
# 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

        slow = head
        fast = head.next

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