题目

给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题（即，只能进行节点交换）。

示例 1：

输入：head = [1,2,3,4]
输出：[2,1,4,3]

示例 2：

输入：head = []
输出：[]

示例 3：

输入：head = [1]
输出：[1]

 题解

用25题的解法k=2

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        int n = 0;
        for (ListNode cur = head; cur != null; cur = cur.next) {
            n++;
        }
        ListNode pre = null;
        ListNode cur = head;
        ListNode dummy = new ListNode(0,head);
        ListNode p0 = dummy;
        while (n >= 2) {
            for (int i = 0; i < 2; i++) {
                ListNode nxt = cur.next;
                cur.next = pre;
                pre = cur;
                cur = nxt;
            }
            ListNode nxt = p0.next;
            p0.next.next = cur;
            p0.next = pre;
            p0 = nxt;
            n -= 2;
        }
        return dummy.next;
    }
}

或者直接交换

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode dummy = new ListNode(0,head);
        ListNode p0 = dummy;
        ListNode p1 = head;
        while (p1 != null && p1.next != null) {
            ListNode p2 = p1.next;
            ListNode p3 = p2.next;

            p0.next = p2;//0->2
            p2.next = p1;//2->1
            p1.next = p3;//1->3

            p0 = p1;//下一个循环0在1的位置，1在3的位置
            p1 = p3;
        }
        return dummy.next;
    }
}