142. Linked List Cycle II

MediumLeetCode

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

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 (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

Example 1

Input: head = [3,2,0,-4], pos = 1

Output: tail connects to node index 1

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: tail connects to node index 0

Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3

Input: head = [1], pos = -1

Output: no cycle

Explanation: There is no cycle in the linked list.

Constraints

• The number of the nodes in the list is in the range [0, 104]
• -105 <= Node.val <= 105
• pos is -1 or a valid index in the linked-list

How to solve the problem

• There is a fact we need to know in advance: when slow mmeets fast pointer, head meets slow at the node where the cycle begins.

Code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        slow = head
        fast = head
        while fast and fast.next: # when slow and fast.next is not None
            slow = slow.next # slow move 1 step
            fast = fast.next.next # fast move 2 step
            if slow is fast: # when slow and fast meet each other
                while slow is not head: # if index1(start from head) and slow(move 1 step everytime) is not meet, then keep moving
                    slow = slow.next
                    head = head.next
                return slow
        return None

Complexity

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