328.Odd Even Linked List

tags: Medium,Linked List

328. Odd Even Linked List

題目描述

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.

範例

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

解答

Javascript

function oddEvenList(head) {
  if (head === null) return null;
  let odd = head;
  let even = head.next;
  const evenHead = even;
  while (even && even.next) {
    odd.next = even.next;
    odd = odd.next;
    even.next = odd.next;
    even = even.next;
  }

  odd.next = evenHead;

  return head;
}

Marsgoat Dec 6, 2022

Python

# 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:
            return head

        odd, even = head, head.next
        evenHead = even

        while even and even.next:
            odd.next = odd.next.next
            even.next = even.next.next
            odd = odd.next
            even = even.next

        odd.next = evenHead
        return head

Kobe Dec 6, 2022

C#

public class Solution {
    public ListNode OddEvenList(ListNode head) {
        if (head == null || head.next == null) return head;

        ListNode odd = head;
        ListNode even = head.next;
        ListNode evenHead = even;

        while (even != null && even.next != null) {
            odd = odd.next = even.next;
            even = even.next = even.next.next;
        }
        
        odd.next = evenHead;

        return head;
    }
}

Jim Dec 6, 2022

Reference

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