LeetCode24 Swap Nodes in Pairs

题意：

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.(Easy)

分析：

链表题，画个图就比较清晰了，就是每两个进行一下翻转，注意翻转部分头尾指针的指向即可，头指针会变，所以用下dummy node。

代码：

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode* swapPairs(ListNode* head) {
         ListNode dummy();
         dummy.next = head;
         head = &dummy;
         while (head -> next != nullptr && head -> next -> next != nullptr) {
             ListNode* temp1 = head -> next;
             head -> next = head -> next -> next;
             ListNode* temp2 = head -> next -> next;
             head -> next -> next = temp1;
             temp1 -> next = temp2;
             head = head -> next -> next;
         }
         return dummy.next;
     }
 };