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.(头二个节点已经事先处理)找出要换位置的两个节点的前驱节点，设为tempHead，设要更换的两个节点为p,q
2.那么直接p节点拿出，tempHead->next=q;p->next=NULL;
3.然后就是普通把p几点插入到q节点后面即可。p->next=q->next;q->next=p;
4.最后把暂时头部节点后移两个位置，进行下一次更换。
图解如下：
代码：
class Solution { public: ListNode* swapPairs(ListNode* head) { ListNode* res; if(head==NULL) return NULL; if(head->next==NULL) return head; //交换第一,二个节点 ListNode* first=head->next; head->next=first->next; first->next=head; ListNode* tempHead=first->next; ListNode* p; ListNode* q; while (tempHead) { if(tempHead->next!=NULL&&tempHead->next->next!=NULL){ p=tempHead->next; q=tempHead->next->next; } else return first; tempHead->next=q; p->next=NULL;//中间节点插入即可 p->next=q->next; q->next=p; tempHead=q->next; } return first; } };
原文：http://www.cnblogs.com/fightformylife/p/4108142.html