Problem description:

Given 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. Note that pos is not passed as a parameter.

Notice that you should not modify the linked list.

Example 1:

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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:


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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.

Solution:

  1. Check if there’s a cycle. If no, return None. If yes, step 2
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L1: distance between the head point and entry point, H
L2: distance from E to X
C: cycle length
                  _____
                 /     \
head_____H______E       \
                \       /
                 X_____/

Since fast = 2 * slow. When they first meet at X.

slow pointer traveled when encounter is L1 + L2

fast pointer traveled when encounter is L1 + L2 + n * C

Because the total distance the fast pointer traveled is twice as the slow pointer, Thus:

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2 * (L1+L2) = L1 + L2 + n * C => L1 + L2 = n * C => L1 = (n - 1) C + (C - L2)*

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

class Solution:
    def detectCycle(self, head: ListNode) -> ListNode:
        if not head:
            return head
        slow, fast = head, head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next
            if fast == slow:
                slow = head
                while fast != slow:
                    fast = fast.next
                    slow = slow.next
                return fast
        return None

time complexity: $O(n)$
space complexity: $O(1)$
reference:
https://leetcode.com/problems/linked-list-cycle-ii/discuss/44781/Concise-O(n)-solution-by-using-C%2B%2B-with-Detailed-Alogrithm-Description
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