leetcode 【 Swap Nodes in Pairs 】python 实现

题目：

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.

代码：oj 测试通过 Runtime: 42 ms

 # Definition for singly-linked list.
 # class ListNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.next = None

 class Solution:
     # @param a ListNode
     # @return a ListNode
     def swapPairs(self, head):
         if head is None or head.next is None:
             return head

         dummyhead = ListNode(0)
         dummyhead.next = head

         pre = dummyhead
         curr = head
         while curr is not None and curr.next is not None:
             tmp = curr.next
             curr.next = tmp.next
             tmp.next = pre.next
             pre.next = tmp
             pre = curr
             curr = curr.next
         return dummyhead.next

思路：

基本的链表操作。

需要注意的是while循环的判断条件：先判断curr不为空，再判断curr.next不为空。

这种and判断条件具有短路功能，如果curr为空就不会进行下一个判断了，因此是安全的