Linked List Cycle II
The key idea
Use Floyd's two-pointer trick. After
slow and fast first meet inside the cycle, the distance from the list head to the cycle entrance equals the distance from the meeting point to the entrance. So reset one pointer to head, advance both one step at a time, and the node where they meet again is the cycle's start.Problem
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.
Constraints
- The number of the nodes in the list is in the range
[0, 10^4]. -10^5 <= Node.val <= 10^5posis-1or a valid index in the linked-list.
Examples
Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Input: head = [1], pos = -1
Output: no cycle
Complexity
Time: O(n) Space: O(1)
See the full solution
- ✓Full worked approach
- ✓Reference code in 5 languages
- ✓Problem-solving tips
- ✓Step-by-step animated visualization