Linked List Cycle II

head berilgan, bog'langan ro'yxatning head holda, tsikl boshlangan tugunni qaytaring. Agar tsikl bo'lmasa, nullni qaytaring.

Agar ro'yxatda keyingi ko'rsatkichni doimiy ravishda kuzatib borish orqali yana erishish mumkin bo'lgan ba'zi tugun bo'lsa, bog'langan ro'yxatda tsikl mavjud. Ichkarida, pos tailning keyingi ko'rsatkichi ulangan tugun indeksini belgilash uchun ishlatiladi (0-indekslangan). Agar tsikl bo'lmasa -1 bo'ladi.

E'tibor bering, pos parametr sifatida o'tkazilmaydi.

Bog'langan ro'yxatni o'zgartirmang.

Example 1:

Linked List Cycle

Input: head = [3,2,0,-4], pos = 1 Output: tail 1-tugun indexiga ulanadi Explanation: Bog'langan ro'yxatda tail ikkinchi tugunga ulanadigan tsikl mavjud.

Example 2:

Linked List Cycle

Input: head = [1,2], pos = 0 Output: tail 0-tugun indexiga ulanadi Explanation: Bog'langan ro'yxatda tail birinchi tugunga ulanadigan tsikl mavjud.

Example 3:

Linked List Cycle

Input: head = [1], pos = -1 Output: tsikl yo'q Explanation: Bog'langan ro'yxatda tsikl mavjud emas.

Constraints:

• Ro'yxatdagi tugunlar soni [0, 104] oralig'ida.

• -105 <= Node.val <= 105

• Bog'langan ro'yxatdagi pos -1 yoki valid indeks.

Follow up: O(1) (ya'ni doimiy) xotiradan foydalanib, uni hal qila olasizmi?

My Solution

func detectCycle(head *ListNode) *ListNode {
	slow, fast := head, head
	hasCycle := false

	for fast != nil && fast.Next != nil {
		fast = fast.Next.Next
		slow = slow.Next
		if slow == fast {
			hasCycle = true
			break
		}
	}

	if !hasCycle {
		return nil
	}

    for slow != head {
        slow = slow.Next
        head = head.Next
    }

	return head
}

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