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.
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.
Approach
Links
GeeksforGeeks
ProgramCreek
YouTube
Examples
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).
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.
Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.
Solutions
// Definition for singly-linked list.
class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
/**
* Time complexity : O(n). We visit each of the nn elements in the list at
* most once. Adding a node to the hash table costs only O(1)O(1) time.
* Space complexity : O(n). The space depends on the number of elements
* added to the hash table, which contains at most nn elements.
*/
public boolean hasCycle(ListNode head) {
Set<ListNode> nodesSeen = new HashSet<>();
while (head != null) {
if (nodesSeen.contains(head)) {
return true;
} else {
nodesSeen.add(head);
}
head = head.next;
}
return false;
}
/**
* Time complexity : O(N), If list has no cycle. O(N+K), If list has a cycle.
* Space complexity : O(1). We only use two nodes (slow and fast).
*/
public class Solution {
public boolean hasCycle(ListNode head) {
if (head == null || head.next == null) {
return false;
}
ListNode slowptr = head;
ListNode fastptr = head.next;
while(slowptr != fastptr) {
if(fastptr == null || fastptr.next == null) {
return false;
}
slowptr = slowptr.next;
fastptr = fastptr.next.next;
}
return true;
}
}
Follow up
Can you solve it using O(1) (i.e. constant) memory?
