题解 | 链表的奇偶重排

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public 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;
        }
        // write code here
        ListNode cur =  head;
        ListNode oushu = new ListNode(0);
        ListNode p = oushu;
        ListNode jishu_next;

        while(cur!=null && cur.next!=null){
            jishu_next = cur.next.next;
            p.next = cur.next;
            cur.next = jishu_next;
            cur = cur.next;
            p = p.next;
            p.next = null;
        }
        // 找到head的最后一个节点
        ListNode m = head;
        while(m.next!=null){
            m = m.next;
        }
        m.next = oushu.next;
        return head;
    }
}

现将奇数链表和偶数链表分开，然后将偶数链表接到奇数链表后面