标签：
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
1 # Definition for singly-linked list. 2 # class ListNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution: 8 # @param a ListNode 9 # @return a ListNode 10 def swapPairs(self, head): 11 if head is None or head.next is None: 12 return head 13 14 dummyhead = ListNode(0) 15 dummyhead.next = head 16 17 pre = dummyhead 18 curr = head 19 while curr is not None and curr.next is not None: 20 tmp = curr.next 21 curr.next = tmp.next 22 tmp.next = pre.next 23 pre.next = tmp 24 pre = curr 25 curr = curr.next 26 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
leetcode 【 Swap Nodes in Pairs 】python 实现
标签：
原文地址：http://www.cnblogs.com/xbf9xbf/p/4214359.html
