[Leetcode]Swap Nodes in Pairs

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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.

这道题有点意思。主要是记住交换后上一个的父节点还没改变呢，所以用递归来做是比较容易处理的。

 1 class Solution {
 2 public:
 3     ListNode *swapPairs(ListNode *head) {
 4         if(head==NULL || head->next==NULL)
 5             return head;
 6         ListNode *temp = head->next;
 7         ListNode *forward = temp->next;
 8         head->next->next = head;
 9         head->next=swapPairs(forward);
10         return temp;
11     }
12 };

[Leetcode]Swap Nodes in Pairs

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原文地址：http://www.cnblogs.com/desp/p/4345793.html