141. Linked List Cycle

Description

See: https://leetcode.com/problems/linked-list-cycle/

Solution

The strategy is to use two pointers, a walker and a runner (slow and fast). The walker (slow) moves forward one node at a time. The runner (fast) moves ahead 2 nodes at a time. So if there is a cycle, the runner will eventually lap the walker and the walker and runner will be at the same node. Otherwise the runner will reach the end of the linked list first.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        if not head or not head.next:
            return False
        walker = head
        runner = head
        while runner and runner.next:
            walker = walker.next
            runner = runner.next.next
            if walker == runner:
                return True
        return False
    

Time Complexity: O(N).

Space complexity O(1)