【Leetcode】之Swap Nodes in Pairs

一.问题描述

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.

二.我的解题思路

这种链表的题一向不难解决，在纸上写写画画往往就知道该怎么求解。我是每两个节点进行一次操作，设立了四个指针。这四个指针分别指向，当前节点上一个节点（pri），当前节点(curr)，下一个节点(next)，下下个节点(tmp)。测试通过的程序如下：

/**
* 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 * curr,* next,* tmp,*res,*pri;
if(head==NULL)
return NULL;
if(head->next==NULL)
return head;
curr=head;next=curr->next;
/*first done*/
tmp=next->next;
next->next=curr;
curr->next=tmp;
res=next;
pri=curr;
curr=curr->next;

while(curr!=NULL){
next=curr->next;
if(next==NULL)
break;
if(next->next!=NULL){
tmp=next->next;
next->next=curr;
curr->next=tmp;
curr=curr->next;
pri->next=next;
pri=next->next;
}
else{
pri->next=next;
next->next=curr;
curr->next=NULL;
}

}
return res;

}
};