【LeetCode】0024 两两交换链表中的节点

https://leetcode-cn.com/problems/swap-nodes-in-pairs/

# -*- coding:utf-8 -*-
# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None
class Solution(object):
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head is None:
            return
        node = head.next
        if node is None:
            return head
        head.next = node.next
        node.next = head
        head.next = self.swapPairs(head.next)
        return node
if __name__ == '__main__':
    n1 = ListNode(1)
    n2 = ListNode(2)
    n3 = ListNode(3)
    n4 = ListNode(4)
    n1.next = n2
    n2.next = n3
    n3.next = n4
    print(Solution().swapPairs(n1))

思路：每次交换两个临近的链表，然后向下递归。边界条件就是剩一个节点（总节点数为奇数），直接返回head即可，正好两两交换（这种的话递归的head就为空）直接返回空即可。