Given a linked list, determine if it has a cycle in it.
To represent a cycle in the given linked list, we use an integer pos
which represents the position (0-indexed) in the linked list where tail connects to. If pos
is -1
, then there is no cycle in the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1 Output: true Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0 Output: true Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1 Output: false Explanation: There is no cycle in the linked list.
Follow up:
Can you solve it using O(1) (i.e. constant) memory?
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def hasCycle(self, head: ListNode) -> bool:
slow = fast = head
while fast and fast.next:
slow, fast = slow.next, fast.next.next
if slow == fast:
return True
return False
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
return true;
}
}
return false;
}
}