Swap Nodes in Pairs

1、这种倒置的问题一半都要重新再选取一个listnode，然后再加上辅助的指针，cur，ret；
2、两个结点两个结点的倒置，间隔的方法为next=ret->next->next;
题目：
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.
思路：
1、选取辅助的listnode进行计算，两个结点两个结点的倒置；
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 *cur=helper;
        ListNode *ret=head;

        while(ret && ret->next)
        {
            ListNode *next=ret->next->next;
            cur->next=ret->next;
            cur=cur->next;
            cur->next=ret;
            ret->next=NULL;
            cur=ret;
            ret=next;
        }
        if(ret) cur->next=ret;
        return helper->next;
    }
};