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：题目不难，假设交换p与q（p在前）需要知道p前面的指针，因此需要三个指针，注意交换头部，以及链表长度为奇数的情况。代码如下：

/**
* 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 || head->next == NULL)//空或者长度为1
return head;
ListNode *p, *q, *pre;
pre = p = head;
q = p->next;
while(q){
if(p == head){//交换头
p->next = q->next;
q->next = p;
head = q;
}
else{
pre->next = q;
p->next = q->next;
q->next = p;
}
//移动
if(p->next){//没有到结尾
pre = p;
p = p->next;
q = p->next;
}
else
break;
}
return head;
}
};