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)
            return null;
        if (head.next == null)
            return head;
        if (head.next.next == null)
            return head;
        ListNode pR, pJS;
        pR = pJS = head;
        ListNode pHOS, pOS;
        pHOS = pOS = head.next;
        int nCount = 0;
        while (head != null) {
            nCount++;
            if (nCount % 2 == 1) {
                ListNode ln = head.next;
                pJS.next = head;
                head = ln;
                pJS = pJS.next;
                pJS.next = null;
            } else {
                ListNode ln = head.next;
                pOS.next = head;
                head = ln;
                pOS = pOS.next;
                pOS.next = null;
            }
        }
        pJS.next = pHOS;
        return pR;
    }
}