❤️ I love the algorithm used here!
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.
true if there is a cycle in the linked list. Otherwise, return
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 = , pos = -1 Output: false Explanation: There is no cycle in the linked list.
- The number of the nodes in the list is in the range
-105 <= Node.val <= 105
-1or a valid index in the linked-list.
Follow up: Can you solve it using
O(1) (i.e. constant) memory?
This uses Floyd's Tortoise and Hare algorithm, the idea is we have tortoise (slow) and hare (fast) pointers.
There is a neat mathematical proof that states if there is a cycle in a list then it must repeat a value twice.
# 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 """ # We can do this with 2 pointers # fast is "fast" as in it moves 2 steps fast = head # slow is "slow" as in it moves 1 step slow = head # since fast moves faster we expect it to end first # so we loop based on it while fast: # if fasts next or next.next is empty # we have reached the end and there is no cycle! if not fast.next or not fast.next.next: break # increment slow and fast slow = slow.next fast = fast.next.next # if slow == fast: return True return False