Swap Nodes in Pairs（链表操作）

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;
}
};