141. Linked List Cycle
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).
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.
Example 3:
Input: head = [1], pos = -1 Output: false 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.
Follow up: Can you solve it using O(1)
(i.e. constant) memory?
Solutions
Solution 1: Hash Table
We can traverse the linked list and use a hash table $s$ to record each node. When a node appears for the second time, it indicates that there is a cycle, and we directly return true
. Otherwise, when the linked list traversal ends, we return false
.
The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes in the linked list.
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Solution 2: Fast and Slow Pointers
We define two pointers, $fast$ and $slow$, both initially pointing to $head$.
The fast pointer moves two steps at a time, and the slow pointer moves one step at a time, in a continuous loop. When the fast and slow pointers meet, it indicates that there is a cycle in the linked list. If the loop ends without the pointers meeting, it indicates that there is no cycle in the linked list.
The time complexity is $O(n)$, and the space complexity is $O(1)$, where $n$ is the number of nodes in the linked list.
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