328. Odd Even Linked List

Medium (Trung bình) Python 🔗 Xem trên LeetCode

📋 Đề Bài

Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.

The first node is considered odd, and the second node is even, and so on.

Note that the relative order inside both the even and odd groups should remain as it was in the input.

You must solve the problem in O(1) extra space complexity and O(n) time complexity.

 

Example 1:

Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]

Example 2:

Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

 

Constraints:

  • The number of nodes in the linked list is in the range [0, 104].
  • -106 <= Node.val <= 106

🧠 Thuật Toán & Kỹ Thuật

Linked List (Danh sách liên kết)
⏱️ Thời gian O(n)
💾 Không gian O(n)

💻 Lời Giải

Python 0328-odd-even-linked-list.py
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def oddEvenList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head or not head.next or not head.next.next:
            return head
            
        odd, even = head, head.next
        headEven = even
        # 1 2 3 4 5
        # odd: 1->3->5, even: 2->4
        while even and even.next:
            odd.next = odd.next.next
            even.next = even.next.next
            odd = odd.next
            even = even.next
        
        odd.next = headEven
        return head