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:
![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist.png)
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:
![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test2.png)
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:
![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test3.png)
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?
Solution(javascript)
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {boolean}
*/
// var hasCycle = function(head) {
// let visited = new Set();
// while (head !== null) {
// if (visited.has(head)) return true;
// visited.add(head);
// head = head.next;
// }
// return false;
// // Time O(n)
// // Space O(n)
// };
var hasCycle = function(head) {
// Floyd's Tortoise And Hare Algorithm
let fast = head;
let slow = head;
while (fast !== null && fast.next !== null) {
slow = slow.next;
fast = fast.next.next;
if (fast == slow) return true;
}
return false;
// Time O(n)
// Space O(1)
};