题解 | #链表的奇偶重排#

using System;
using System.Collections.Generic;

/*
public class ListNode
{
	public int val;
	public ListNode next;

	public ListNode (int x)
	{
		val = x;
	}
}
*/

class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    public ListNode oddEvenList (ListNode head) {
        if(head == null) return null;
        ListNode odd = head;
        ListNode even = head.next;
        ListNode cur = odd;
        ListNode nex;
        bool isOdd = true;
        while(cur != null && cur.next != null){
            nex = cur.next;
            cur.next= nex.next;
            cur = nex;
            isOdd = !isOdd;
            if(isOdd) odd = cur;
        }
        odd.next = even;
        return head;
    }
}