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.

思路：记录两个工作指针 p和q；

          记录原来链表的当前位置cur；

         记录新链表的尾指针curlist；

        一对一对进行交换即可！

/** * 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)         return NULL;       if(head->next == NULL)         return head;       ListNode * q = head;       ListNode * p = q->next;       ListNode * cur = head->next->next;//旧节点未加入初始节点       ListNode * curlist = head;//新链表的尾节点       /*链表头交换*/       q ->next = p->next;       p->next = q;       head = p;//当前链表的头节点       curlist = q;//当前链表的尾节点       /*链表 恢复*/       if(cur == NULL )         return head;       if(cur->next == NULL)       {            curlist->next = cur;            return head;       }       while(cur != NULL && cur->next != NULL)       {           q = cur;//原来链表的位置           p = cur->next;           cur = cur->next->next;           q ->next = p->next;           p->next = q;           curlist->next = p;           curlist = q;           if(cur == NULL)              return head;            if(cur->next == NULL)            {                curlist->next = cur;                return head;            }       }    }};