Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

思路：題目比較簡單，會連結串列反轉的都可以做，思路也差不多，不多說，上程式碼。

/**
 * 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 firstHead = new ListNode(0);
	        firstHead.next = head;
	        ListNode pre = firstHead;//定義一個頭結點，這樣所有的操作都相同
	        ListNode next = null;
	        while(head != null && head.next != null){
	        	//A-B-C-D交換BC,pre=A;B=head;C=next
	            next = head.next;//儲存交換的變數C
	            
	            head.next = next.next;//將B指向B的指標指向D
	            pre.next = next;//將A指向B的指標指向C
	            next.next = head;//將C指向D的指標指向B,完成交換，順序變為A-C-B-D
	            
	            //為下一迴圈準備變數
	            pre = head;//將pre變為B
	            head = head.next;//將head指向D
	            
	        }
	        return firstHead.next;
	    }
}