(链表 递归) leetcode 24. Swap Nodes in Pairs

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

You may not modify the values in the list's nodes, only nodes itself may be changed.

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

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中文大意是两两交换其中相邻的结点。有非递归和递归的解法，在非递归的解法上，要画下图，辅助理解。

C++代码：
非递归：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode *dummy = new ListNode(-1),*cur = dummy;
        cur->next = head;
        while(cur->next && cur->next->next){
            ListNode *t = cur->next->next;
            cur->next->next = t->next;
            t->next = cur->next;
            cur->next = t;
            cur =cur->next->next;
        }
        return dummy->next;
    }
};

递归：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if(!head || !head -> next){
            return head;
        }
        ListNode *t = head->next;
        head->next = swapPairs(head->next->next);
        t->next = head;
        return t;
    }
};

参考博客：http://www.cnblogs.com/grandyang/p/4441680.html