Squares of a Sorted Array

This simple LeetCode problem (Problem #141) demonstrates the power of two pointers to reduce runtime.

Problem:

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:

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:

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 (0-indexed).

Example 3:

Input: head = [1], pos = -1

Output: false

Explanation: There is no cycle in the linked list.

Solution

We can solve this problem easily using a variation of the two-pointers technique, where the second pointer advances faster than the first pointer. We can let the first pointer move 1-index at a time, while the second pointer move 2-indices at a time.

If there is no cycle, the next pointer of the last node will be “None” and the while loop will terminate.

If there is a cycle, at some point during the looping, the faster pointer will be at the node, whose next pointer will be pointing to the node currently indexed by the slower pointer.

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

def hasCycle(head: Optional[ListNode]):
    slow = fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
        if slow == fast:
            return True
    return False

Runtime: Runtime is $O(n)$ and the space complexity is $O(1)$.

Back to Two Pointers Problems

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