Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.
Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
Solution. 遞歸
應(yīng)該很容易想到用遞歸來(lái)分解計(jì)算步驟：
a. 先交換最前面兩個(gè)元素
b. 再算后面的List
c. 將a/b兩步結(jié)果接起來(lái)
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 not head or not head.next:
return head
_head = head.next
head.next = self.swapPairs(_head.next)
_head.next = head
return _head
