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) {
        // write code here
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummyOdd = new ListNode(-1);
        ListNode dummyEven = new ListNode(-1);
        ListNode tailOdd = dummyOdd;
        ListNode tailEven = dummyEven;
        int i = 1;
        ListNode cur = head;
        while (cur != null) {
            if (i % 2 != 0) {
                tailOdd.next = cur;
                tailOdd = cur;
            } else {
                tailEven.next = cur;
                tailEven = cur;
            }
            cur = cur.next;
            i++;
        }
        tailOdd.next = dummyEven.next;
        tailEven.next = null;
        return dummyOdd.next;
    }
}