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
     * ***双指针法
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    public ListNode oddEvenList (ListNode head) {
        if(head == null)
            return head;
        //奇数位指针
        ListNode odd = head;
        //偶数链的头节点为第二个节点
        ListNode evenHead = head.next;
        //偶数位指针
        ListNode even = evenHead;
        //全部节点分离完毕的条件，此时odd指向最后一个奇数位
        while(even != null && even.next != null){
            odd.next = even.next;
            odd = odd.next;
            even.next = odd.next;
            even = even.next;
        }
        //将偶数链表连在奇数链表之后
        odd.next = evenHead;
        return head;
    }
}