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;
}

}

}
};