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]提示：链表中节点的数目在范围 [0, 100] 内0 <= Node.val <= 100

/** * Definition for singly-linked list. * function ListNode(val) { *     this.val = val; *     this.next = null; * } *//** * @param {ListNode} head * @return {ListNode} */var swapPairs = function (head) {  const dummy = new ListNode(0);  dummy.next = head;  let current = dummy;  while (current.next != null && current.next.next != null) {    // 初始化双指针    const first = current.next;    const second = current.next.next;    // 更新双指针和 current 指针    first.next = second.next;    second.next = first;    current.next = second;    // 更新指针    current = current.next.next;  }  return dummy.next;};

class Solution:    def swapPairs(self, head: ListNode) -> ListNode:        """        用递归实现链表相邻互换：        第一个节点的 next 是第三、第四个节点交换的结果，第二个节点的 next 是第一个节点；        第三个节点的 next 是第五、第六个节点交换的结果，第四个节点的 next 是第三个节点；        以此类推        :param ListNode head        :return ListNode        """        # 如果为 None 或 next 为 None，则直接返回        if not head or not head.next:            return head        _next = head.next        head.next = self.swapPairs(_next.next)        _next.next = head        return _next

/** * Definition for singly-linked list. * type ListNode struct { *     Val int *     Next *ListNode * } */func swapPairs(head *ListNode) *ListNode {if head == nil || head.Next == nil {return head}next := head.Nexthead.Next = swapPairs(next.Next) // 剩下的节点递归已经处理好, 拼接到前 2 个节点上next.Next = headreturn next}

/** * Definition for a singly-linked list. * class ListNode { *     public $val = 0; *     public $next = null; *     function __construct($val = 0, $next = null) { *         $this->val = $val; *         $this->next = $next; *     } * } */class Solution{    /**     * @param ListNode $head     * @return ListNode     */    function swapPairs($head)    {        if (!$head || !$head->next) return $head;        /** @var ListNode $next */        $next = $head->next;        $head->next = (new Solution())->swapPairs($next->next); // 递归已经将后面链表处理好, 拼接到前面的元素上        $next->next = $head;        return $next;    }}

class Solution {public:    ListNode* swapPairs(ListNode* head) {        ListNode h, *tail = &h;        while (head && head->next) {            auto p = head, q = head->next;            head = q->next;            q->next = p;            tail->next = q;            tail = p;        }        tail->next = head;        return h.next;    }};