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
# 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
p = dummyhead while p.next is not None and p.next.next is not None:
tmp = p.next.next.next
p.next.next.next = p.next
p.next = p.next.next
p.next.next.next = tmp
p = p.next.next 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
# 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 dummpyhead = ListNode(0)
dummpyhead.next = head p = dummpyhead while p.next is not None and p.next.next is not None:
tmp = p.next
p.next = p.next.next
tmp.next = p.next.next
p.next.next = tmp
p = p.next.next return dummpyhead.next
