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){
            return head;
        }
        ListNode evenHead = head.next;
        ListNode odd = head;
        ListNode even = evenHead;
        while(even!=null&&even.next!=null){
            odd.next = even.next;
            odd = odd.next;
            even.next = odd.next;
            even = even.next;             
        }
        odd.next = evenHead;
        return head;
    }
}

本题的核心在于：理解奇偶节点的特殊位置关系