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：

/**
* 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;
}
};