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：题目要求交换链表的相邻的两个节点，但是不许修改节点的数据，只是移动节点的位置。可以给这个链接加一个头节点然后进行处理，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) {
ListNode *res=new ListNode(0),*pre=res,*cur;
pre->next=head;
cur=head;
while(cur&&cur->next)
{
pre->next=cur->next;
cur->next=cur->next->next;
pre->next->next=cur;
pre=cur;
cur=cur->next;
}
return res->next;

}
};方法二：在discuss上看到有使用递归的方式，可以参考下
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if(head == NULL||head->next==NULL)
return head;
ListNode* next = head->next;
head->next = swapPairs(next->next);
next->next = head;
return next;
}
};