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.
解法：可以使用递归，每次交换前两个，将剩下的链表递归，返回的是新的链表头。
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode swapPairs(ListNode head) { 14 if (head==null || head.next==null) { 15 return head; 16 } 17 ListNode p = head; 18 ListNode begin=null; 19 ListNode temp = p; 20 21 p=p.next; 22 temp.next=p.next; 23 p.next=temp; 24 if (begin==null) { 25 begin = p; 26 } 27 p=p.next; 28 p.next=swapPairs(p.next); 29 return begin; 30 } 31 }
原文：http://www.cnblogs.com/birdhack/p/4001064.html