Linked List Cycle
The key idea
Walk two pointers at different speeds.
slow advances one node per step, fast advances two. If there is a cycle, fast eventually laps slow and they land on the same node. If fast reaches null, the list ends, so there is no cycle.Problem
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 the 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.
Constraints
- The number of 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: true
Input: head = [1,2], pos = 0
Output: true
Input: head = [1], pos = -1
Output: false
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