题目：
Swap Nodes in Pairs
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164ms通过
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 12 if head is None or head.next is None: 13 return head 14 15 dummyhead = ListNode(0) 16 dummyhead.next = head 17 p = dummyhead 18 19 while p.next is not None and p.next.next is not None: 20 tmp = p.next.next.next 21 p.next.next.next = p.next 22 p.next = p.next.next 23 p.next.next.next = tmp 24 p = p.next.next 25 26 27 return dummyhead.next
思路：
1. 建立一个虚表头hummyhead 这样方便操作一些
2. p.next始终指向要交换的下一个元素的位置
3. 先保存p.next.next.next，再移动p，p.next，p.next.next，p.next.next.next：先动p.next.next.next再动其他的。
小白我一开始先动的是p，p.next结果后面的p.next.next就丢了，其他小白别陷入这个误区，高手请略过。
Tips: 动了哪个指针，就把哪个指针上面打个×；添加了哪个指针，就在两个点之间加一根线；画画图就出来了，别光看着不动笔。
又做了一遍 第二次ac的 小失误了
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 dummpyhead = ListNode(0) 15 dummpyhead.next = head 16 17 p = dummpyhead 18 19 while p.next is not None and p.next.next is not None: 20 tmp = p.next 21 p.next = p.next.next 22 tmp.next = p.next.next 23 p.next.next = tmp 24 p = p.next.next 25 26 return dummpyhead.next