一、題目敘述：

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.

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二、解題思路：

easy題，非常簡單，不要想著交換連結串列的指標，交換值就行了。

（1）之後在做完題後，把最基本的例子跑一遍，不要再寫死迴圈了！

三、原始碼：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution 
{
    public ListNode swapPairs(ListNode head) 
    {
     ListNode a = head;
     int temp;
     if (head == null) return null;
     while (a.next != null)
     {
    	 temp = a.next.val;
    	 a.next.val = a.val;
    	 a.val = temp;
    	 if (a.next.next != null)
    		 a = a.next.next;
    	 else 
    		 a = a.next;
    	 
     }
     return  head;
    }
    public void print(ListNode root)     
    {    
      System.out.print(root.val + "\t");    
      if (root.next != null)    
    	  print(root.next);    
    }    
    public static void main (String args[])
    {
    	 ListNode l1 = new ListNode(2);    
         ListNode a = l1;    
         a.next = new ListNode(3);    
         a = a.next;    
         a.next = new ListNode(4);    
         a = a.next;    
         a.next = new ListNode(1); 
         //b.next = new ListNode(4);    
         Solution s = new Solution();    
         s.print(s.swapPairs(l1));    
    }
}