标签：
Swap Nodes in Pairs
问题：
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.
思路：
画画图即可，加入几个first second dummy的变量就顺理成章了
我的代码：
public class Solution { public ListNode swapPairs(ListNode head) { if(head == null) return head; ListNode dummy = new ListNode(-1); dummy.next = head; ListNode pre = dummy; ListNode first = head; ListNode second = head.next; while(first != null && second != null) { ListNode next = second.next; pre.next = second; second.next = first; first.next = next; pre = first; if(next == null) return dummy.next; first = next; second = next.next; } return dummy.next; } }
标签：
原文地址：http://www.cnblogs.com/sunshisonghit/p/4319890.html
