Linked List Cycle

easy LeetCode #141

hash-table linked-list two-pointers

Description

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

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter.

Return true if there is a cycle in the linked list. Otherwise, return false.

Example 1:
Input: head = [3, 2, 0, -4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).
3 -> 2 -> 0 -> -4
^-----------|

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

Example 3:
Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list (pos = -1 means no cycle).

Constraints

The number of nodes in the list is in the range [0, 10^4]
-10^5 <= Node.val <= 10^5
pos is -1 or a valid index in the linked-list

Complexity

Show Complexity

Time: O(n)
Space: O(1)

Hints

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Pattern: Two pointers (Floyd's Cycle Detection / Tortoise and Hare)
Approach: Use two pointers: slow moves 1 step, fast moves 2 steps. If there's a cycle, they will eventually meet. If fast reaches null, there's no cycle.
Complexity: O(n) time with O(1) space using slow/fast pointers

Solutions

Show PY Solution

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


def solution(head: list[int], pos: int) -> bool:
    n = None
    l = None
    for i, h in enumerate(head):
        nn = ListNode(h)
        if n:
            n.next = nn
        if i == pos:
            l = nn
        n = nn
        if i == len(head) - 1:
            n.next = l

    slow = fast = n

    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
        if slow is fast:
            return True

    return False

    # n = None
    # l = None
    # for i, h in enumerate(head):
    #     nn = ListNode(h)
    #     if n:
    #         n.next = nn
    #     if i == pos:
    #         l = nn
    #     n = nn
    #     if i == len(head) - 1:
    #         n.next = l
    #
    # if l is None:
    #     return False
    #
    # d = l
    # while d:
    #     if d.val == n.val:
    #         return True
    #     d = d.next
    #
    # return False