leetcode: Swap Nodes in Pairs

相当于Reverse Nodes in k-Group这道题的简化版，固定k为2，每两个翻转一下，方法和思路也一摸一样。  要更简单一些。空间消耗为常数，时间消耗为O(n)。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        if( head==null )
        {
            return head;
        }
        ListNode res = head;
        int count=0;
        boolean first=true;
        ListNode pre=null;
        ListNode st=head;
        while( head !=null )
        {
            count++;
            if( count==2 )
            {
                ListNode tmp=head.next;
                head.next=st;
                st.next=tmp;
                if( pre != null )
                {
                    pre.next=head;
                }
                if( first == true )
                {
                    res = head;
                    first=false;
                }
                pre = st;
                st = tmp;
                head = tmp;
                count=0;
            }
            else
            {
                head = head.next;
            }
        }
        return res;
    }
}