题解 | #两两交换链表的节点#

利用反转链表的思路加上每 k 个 节点的限制,对整个链表进行递归处理, 得出最终的结果

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 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
#
class Solution:
    def reverse(self, s: ListNode, e: ListNode) -> ListNode:
        t = None
        while s != e:
            n = s.next
            s.next = t
            t = s
            s = n
        return t
    
    def reverseK(self, head: ListNode, k: int) -> ListNode:
        s = head
        for i in range(k):
            if not s:
                return head
            s = s.next
        b = self.reverse(head, s)
        head.next = self.reverseK(s, k)
        return b
    
    
    def swapLinkedPair(self , head: ListNode) -> ListNode:
        # write code here
        return self.reverseK(head, 2)