leetcode 【 Linked List Swap Nodes in Pairs 】 python 实现

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