LeetCode#24 Swap Nodes in Pairs

Problem Definition:

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: 题目要求不能直接交换链表结点的元素值。（虽然跑起来OJ也发现不了）

用递归实现起来是比较简单直观的，无需构造首节点什么的。这里要注意处理奇数个节点的情况。

    # @param {ListNode} head
    # @return {ListNode}
    def swapPairs(self, head):
        return self.recur(head, head.next) if head else None

    def recur(self, p, q):
        if q==None:
            return p
        tmp=q.next
        q.next=p
        p.next=self.recur(tmp, tmp.next) if tmp else None
        return q