题解 | #【冲刺双百】链表的奇偶重排#

思路

1. 每次看两个，隔一个指向一个
2. 注意分类讨论以及链表指向的关系

public class Solution {

    public ListNode oddEvenList (ListNode head) {
        if(head==null||head.next==null) return head;
        ListNode now=head;
        ListNode now2=head.next;
        ListNode head2=now2;
        while(now2.next!=null&&now2.next.next!=null){
            now.next=now2.next;
            now=now.next;
            now2.next=now.next;
            now2=now2.next;
        }
        if(now2.next==null){
            now.next=head2;
        }
        else{
            now.next=now2.next;
            now=now.next;
            now2.next=null;
            now.next=head2;
        }
            return head;
    }
}