leetcode24，交换链表相邻的节点

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.

对于这样链表交换的题目，我一般都有两种解法。

第一种，看链表每个节点带有的数据量，这道题带有只有一个value值，只有一个int值，所以我一开始的想法就是，不真实的交换节点，只是交换里面的数据。

这样的好处是空间复杂度特别省，适合那些每个节点数据很少，要求不高的链表交换。

+ View code

/**
 * 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) {
        if(head == null || head.next == null)
            return head;
        
        ListNode A = head;
        ListNode B = head.next;
        
        int t;
        while(A != null && B != null){
            t = B.val;
            B.val = A.val;
            A.val = t;
            
            if(B.next == null)
                break;
            A = B.next;
            B = B.next.next;
        }
        
        return head;
    }
}