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 (null == head) {
            return null;
        }
        if (null == head.next) {
            return head;
        }
        if (null == head.next.next) {
            return head;
        }
	    // 双指针 向后移动
        ListNode a = head;
        ListNode b = head.next;
	    // 记录第二个节点
        ListNode c = head.next;
        // 链表长度
        int i = 0;
        ListNode k = head;
        while(null != k) {
            i++;
            k = k.next;
        }
        // 长度区分奇偶，结束条件不同
        if (i % 2 == 0) {
            while (null != b.next) {
               a.next = b.next;
               a = b;
               b = b.next;
            }
            a.next = c;
        } else {
            while (null != b) {
              a.next = b.next;
              a = b;
              b = b.next;
            }
            a.next = c;
        }
        return head;
    }
}