#leetcode#24. 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.
没什么难的
就是注意NULL的情况

以及开始的情况

直接新建一个开头

非递归

/**
* 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)
return head;
ListNode *newhead = new ListNode(0);
newhead->next=head;
ListNode *start=newhead;
ListNode *swap1=head;
ListNode *swap2=head->next;
ListNode *end=NULL;
while(swap1!=NULL && swap2!=NULL)
{
if(swap2->next!=NULL)
end=swap2->next;
else
end=NULL;
swap1->next=end;
swap2->next=swap1;
start->next=swap2;
start=swap1;
swap1=end;
if(swap1!=NULL && swap1->next!=NULL)
swap2=swap1->next;
else
swap2=NULL;
}
return newhead->next;
}
};

递归版
好简单

/**
* 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)
return head;
ListNode *newhead=head->next;
head->next=newhead->next;
newhead->next=head;
newhead->next->next=swapPairs(newhead->next->next);
return newhead;
}
};