[Leetcode] Swap nodes in pairs 成对交换结点

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given1->2->3->4, you should return the list as2->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.

题意：成对交换结点。

思路：这题感觉是reverse nodes in k grops的是一个特殊情况。这题更简单一些，只要保证当前结点cur和其后继存在就可以交换。改变表头，所以要new一个。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 *nList=new ListNode(-);

         nList->next=head;

         ListNode *pre=nList;

         ListNode *cur=head;

         while(cur&&cur->next)

         {

             ListNode *temp=cur->next;

             cur->next=temp->next;

             temp->next=pre->next;

             pre->next=temp;

             pre=cur;

             cur=cur->next;

         } 

         return nList->next;

     }

 };

巴特西

[Leetcode] Swap nodes in pairs 成对交换结点

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