328. Odd Even Linked List

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

---

# https://leetcode.com/problems/odd-even-linked-list/description/

'''
1. 아이디어 :
    1) 링크드 리스트를 순회하며, 두번째 인덱스를 Even Head로 만든다.
    odd, even을 번갈아가며 다음 노드를 연결시키고, 마지막 odd의 even head를 연결시킨다.
2. 시간복잡도 :
    1) O(n)
3. 자료구조 :
    1) Linked List
'''
# 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 head==None:
            return None
        odd=head
        even=odd.next
        evenh=even
        
        while odd.next and even.next:
            odd.next=even.next
            odd=even.next
            even.next=odd.next
            even=odd.next
        odd.next=evenh
        return head