Classical Python solution


  • 0
    G
    # Definition for singly-linked list.
    # class ListNode:
    #     def __init__(self, x):
    #         self.val = x
    #         self.next = None
    
    class Solution:
        # @param head, a ListNode
        # @return a boolean
        def hasCycle(self, head):
            if not head:
                return False
    
            slow = fast = head
            while fast and fast.next:
                slow = slow.next
                fast = fast.next.next
                if slow is fast:
                    return True
    
            return False

  • 0
    C

    Nice solution, while the first two checking lines are not necessary.


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