328. Odd Even Linked List

Given a singly linked list, group all odd nodes together followed by
the even nodes. Please note here we are talking about the node number
and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space
complexity and O(nodes) time complexity.

Example 1:

Input: 1->2->3->4->5->NULL Output: 1->3->5->2->4->NULL Example 2:

Input: 2->1->3->5->6->4->7->NULL Output: 2->3->6->7->1->5->4->NULL
Note:

The relative order inside both the even and odd groups should remain
as it was in the input. The first node is considered odd, the second
node even and so on ...

思路

用當前鏈表生成兩個鏈表, 分別由當前鏈表的奇數項和偶數項組成,而後收尾相連造成新鏈表
注意遍歷鏈表的退出條件應該是偶數項 even!=null && even.next!=null

複雜度

時間O(n) 空間O(1)

class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode rightHead = head.next;
        ListNode odd = head;
        ListNode even = rightHead;
        while (even!= null && even.next!= null) {
            odd.next = odd.next.next;
            odd = odd.next;
            even.next = even.next.next;
            even = even.next;
        }
        odd.next = rightHead;
        return head;
    }
}