Leetcode刷题记——24. Swap Nodes in Pairs(交换成对结点)

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一、题目叙述：

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、解题思路：

easy题，非常简单，不要想着交换链表的指针，交换值就行了。

（1）之后在做完题后，把最基本的例子跑一遍，不要再写死循环了！

三、源码：

/** * 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));        }}

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