142.Linked List Cycle II

tags: Medium,Linked List,Two Pointers,Hash Table

142. Linked List Cycle II

題目描述

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

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 (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

範例

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
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: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
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?

解答

Python

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head:
            return

        def getIntersect(head):
            slow, fast = head, head

            while fast and fast.next:
                slow = slow.next
                fast = fast.next.next

                if slow == fast:
                    return slow

            return
        
        intersect = getIntersect(head)
        if not intersect:
            return

        ptr1, ptr2 = head, intersect
        while ptr1 != ptr2:
            ptr1 = ptr1.next
            ptr2 = ptr2.next

        return ptr1

Ron ChenThr, Mar 9, 2023

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        fast = slow = head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
            if fast == slow:
                while head != slow:
                    head = head.next
                    slow = slow.next
                return head
        return None

Yen-Chi ChenThu, Mar 9, 2023

C++

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        auto fast = head, slow = head;
        while (fast && fast->next) {
            fast = fast->next->next;
            slow = slow->next;
            if (fast == slow) {
                while (head != slow) {
                    head = head->next;
                    slow = slow->next;
                }
                return head;
            }
        }
        return NULL;
    }
};

Yen-Chi ChenThu, Mar 9, 2023

Javascript

function detectCycle(head) {
  if (head === null) return null;
  let tortoise = head;
  let hare = head;

  do {
    if (hare.next === null || hare.next.next === null) return null;
    tortoise = tortoise.next;
    hare = hare.next.next;
  } while (tortoise !== hare);

  hare = head;

  while (tortoise !== hare) {
    hare = hare.next;
    tortoise = tortoise.next;
  }

  return hare;
}

MarsgoatThu, Mar 9, 2023

Reference

先從這題做起
https://leetcode.com/problems/linked-list-cycle/

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