141. Linked List Cycle (HashTable+ Two Pointers) 20.3.27 Easy

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?

Solution no.1: -------------------------------------------------- HashTable
------------------------------------------------- O(n) T, O(n) S
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if not head:
            return False
        HashTable = {}
        while head:
            if head in HashTable:
                return True
            HashTable[head] = head.val
            head = head.next
        return False
        
Solution no.2: ----------------------------------- Two Pointers ----------------------- https://www.youtube.com/watch?v=9SD2ccDW5CY
------------------------------------------------- O(n) T, O(1) S
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def hasCycle(self, head):
        if not head:
            return False
        slow, fast = head, head.next        # Create two pointers: fast and slow, as long as there is a cycle, 
                                            # these two pointers will meet each other
        while fast and fast.next:           # Must be while fast and fast.next instead of while fast:
                                            # Since below, we use fast = fast.next.next, Nonetype node has no next attribute
                                            # if fast and fast.next change into None, then the link is a straight link witout cycle
                                            # then the while loop is broken and we return False
            if slow == fast:                # if they meet each other, then there exists a cycle and we can return True
                return True                 # Why they will definitely meet each other in a cycled linked list ?
                                            # Because the fast pointer will also be advanced than the slow pointer each step per time,
                                            # so in this way, the fast pointer will  catch up with the slow pointer from behind,
                                            # as long as the length of per cycle is not infinte
            slow = slow.next
            fast = fast.next.next
        return False
        """
        :type head: ListNode
        :rtype: bool
        """


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