[LEET CODE] 328. Odd Even Linked List

# Problem

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

 

# My Answer

# 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: ListNode) -> ListNode:
        if head is None:
            return head
        temp = even = head.next # 짝수 헤드 temp
        odd = head
        while odd.next and odd.next.next:
                odd.next = odd.next.next
                odd = odd.next
                even.next = odd.next 
                even = even.next
        odd.next = temp    
        return head

 

# Solution 1 - 반복 구조로 홀짝 노드 처리

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        # 예외 처리
        if head is None:
            return None

        odd = head
        even = head.next
        even_head = head.next # 짝수의 헤드

        # 반복하면서 홀짝 노드 처리
        while even and even.next:
            odd.next, even.next = odd.next.next, even.next.next
            odd, even = odd.next, even.next

        # 홀수 노드의 마지막을 짝수 헤드로 연결
        odd.next = even_head
        return head