leetcode：Swap Nodes in Pairs

Given a linked list, swap every two adjacent（相邻的） 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.

分析：题意为给予一个链表，交换相邻的两个节点然后返回头结点。要求算法空间复杂度为O(1).

思路：新建一个链表每次插入temp->next后再插入temp，注意要判断若最后只剩下一个节点则不需要交换，每次交换了节点要把尾节点的下一个指向空。

代码：

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/**  * 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) {         if(head==NULL || head->next==NULL) return head;         ListNode *helper=new ListNode(0);         ListNode *temp=head;         ListNode *cur=helper;         while(temp  && temp->next)         {             ListNode *next=temp->next->next;             cur->next=temp->next;             cur=cur->next;             cur->next=temp;             cur=cur->next;             cur->next=NULL;             temp=next;         }         if(temp) cur->next=temp;         return helper->next;     } };