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141. Linked List Cycle.md

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1. Description

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.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Constraints:

  • The number of the nodes in the list is in the range [0, 104].
  • -105 <= Node.val <= 105
  • pos is -1 or a valid index in the linked-list.

Follow up: Can you solve it using O(1) (i.e. constant) memory?

2. Solutions

Solution 1: Language: C Fast and Slow Pointers

If here is a cycle in the linked list, the fast pointer would catching the slow one from behind. We need to make sure that fast != NULL and fast->next != NULL, cause we put fast = fast->next->next in the last iteration.

  • Thursday, 29 July, 2021
  • Time Complexity: $O(n)$
  • Space Complexity: $O(1)$
  • Runtime: 16 ms, faster than 19.68% of C online submissions for Linked List Cycle.
  • Memory Usage: 7.8 MB, less than 96.01% of C online submissions for Linked List Cycle.
bool hasCycle(struct ListNode *head) {
    if (head == NULL) return false;
    struct ListNode *fast = head->next;
    struct ListNode *slow = head;
    while (fast != slow) {
        if (fast == NULL || fast->next == NULL) {
            return false;
        }
        fast = fast->next->next;
        slow = slow->next;
    }
    return true;
}