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

應該很容易想到用遞歸來分解計算步驟：

a. 先交換最前面兩個元素
b. 再算后面的List
c. 將a/b兩步結果接起來

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