ｐｒｏｂｌｅｍ：

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.

在单链表中。每两个结点交换一下位置。单个的不交换

ｔｈｉｎｋｉｎｇ：

（１）这道题在不新建结点的情况下。指向关系复杂。慢慢分析

（２）ｈｅａｄ是头指针，指向第一个有效结点！

！

！

ｃｏｄｅ：

/**

 * 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) {

        ListNode *index = head;

        ListNode *pre = NULL;

        ListNode *tmp = NULL;

        ListNode *modify = head;;

        int i=0;

        if(head==NULL||head->next==NULL)

            return head;

        while((index!=NULL)&&(index->next!=NULL))

        {

            i++;

            pre=index;

            if(i==1)

            {

                head=pre->next;

                index = index->next;

                tmp = index->next;

                pre->next = tmp;

                index->next = pre;

                index=tmp;

                modify=pre;

            }

            else

            {

                index = index->next;

                tmp = index->next;

                modify->next=pre->next;

                pre->next = tmp;

                index->next = pre;

                index=tmp;

                modify=pre;

            }

        }

        return head;

    }

};