Linked List Cycle

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

Algorithm:

The classic fast & slow pointer problem

Just like the clock has 3 hands with different speed.

If there's a cycle, they will meet eventually.

Solutions:

/**
 * 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, fast = head;
        while (fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
            
            if (slow == fast) return true;
        }
        
        return false;
        
    }
}

Performance:

More intuitively, is to save each node reference in a HashSet (Sorted unique elements hash table).
Return true if found a duplicate. But this method takes more time(Around 10 ms) and need extra space.

Solution 2:

public class Solution {
    public boolean hasCycle(ListNode head) {
        Set<ListNode> mySet = new HashSet<ListNode>();
        while(head != null){
            if(mySet.contains(head)){
                return true;
            }
            else{
                mySet.add(head);
                head = head.next;
            }
        }
        return false;
    }
}