24. 两两交换链表中的节点 【递归】

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）。

输入：head = [1,2,3,4]
输出：[2,1,4,3]

 

思路：递归

from typing import Optional
# 创建链表
def create_linked_list(lst):
    if not lst:
        return None

    head = ListNode(lst[0])
    current = head

    for val in lst[1:]:
        current.next = ListNode(val)
        current = current.next

    return head

# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class Solution:
    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if head is None or head.next is None:
            return head
        t = self.swapPairs(head.next.next)
        p = head.next
        p.next = head
        head.next = t
        return p

if __name__ == '__main__':
    s = Solution()
    lst = [1, 2, 3, 4]
    head = create_linked_list(lst)
    res = s.swapPairs(head)
    print(res)