【leetcode】24. Swap Nodes in Pairs

题目描述：

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

解题分析：

解题思路很简单，就是先两两扫描，然后调换顺序即可。这道题的难点在于每个节点的next域指向问题，所以解这样的题最好可以在纸上写一下交换的步骤就不难实现

具体代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public static ListNode swapPairs(ListNode head) {
          if(head==null)
              return null;
          if(head.next==null){
              return head;
          }
          if(head.next.next==null){
              ListNode current=head.next;
              current.next=head;
              head=head.next;
              current.next.next=null;
              return head;
          }
          ListNode pre=null;
          ListNode current=head.next;
          head.next=current.next;
          current.next=head;
          head=current;
          current=head.next;
          pre=current;
          current=pre.next;
          while(current!=null && current.next!=null){
              pre.next=current.next;
              current.next=current.next.next;
              pre.next.next=current;
              pre=current;
              current=pre.next;
          }
          return head;  
     }
    
}