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) {
        // write code here
        if(head==null||head.next==null) return head;
        ListNode t = head;
        //存放最后链表的数
        List<ListNode> jiList = new List<ListNode>();
        List<ListNode> ouList = new List<ListNode>();
        jiList.Add(t);
        //是链表中的第几个结点
        int count = 1;
        while (t.next!= null) {
            t = t.next;
            ++count;
            if (count % 2 == 0) {
                ouList.Add(t);
            } else {
                jiList.Add(t);
            }
        }
        ListNode res = new ListNode(-1);//辅助头结点
        ListNode tail = res;
        for (int i = 0; i < jiList.Count; i++) {
            tail.next = jiList[i];
            tail = jiList[i];
        }
        for (int i = 0; i < ouList.Count; i++) {
            tail.next = ouList[i];
            tail = ouList[i];
        }
        tail.next = null;
        return res.next;
    }
}