Leetcode 24. Swap Nodes in Pairs

Description: Given a linked list, swap every two adjacent nodes and return its head. You may not modify the values in the list's nodes. Only nodes itself may be changed.

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

想法: 如果链表是空，或只有一个元素，直接返回head. 两个元素及以上时，相邻两个元素交换。以[1,2,3,4]为例，交换完[1,2] -> [2,1], 后面的[3,4]可以看做一个新的链表，进行同样的操作，所以可以用递归处理。这时候我们将问题简化，只考虑[1,2,3]，那么我们会怎么做呢？

p = head.next #2
f = p.next #3
p.next = head
head.next = f # f represents the elements followed p 
return p

我们把这段写进递归函数中，将f替换就可以了：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        def swap(l):
            if not l: return l
            if not l.next: return l
            p = l.next
            f = p.next
            p.next = l
            l.next = swap(f)
            return p
        return swap(head)

日期：2020-11-15 (It's a good start to be accepted by one submission, fighting! 冬天来了，居然还是很暖和)