2.28SwapNodesinPairs

Notes:
	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.
	Solution: 1. Iterative solution with constant space.
	2. Recursive solution with O(n) space (for practice).
	*/
	/**
	* 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) {
        return swapPairs_1(head);
    }
    
    ListNode *swapPairs_1(ListNode *head) {
        ListNode dummy(0), *cur = &dummy;
        cur->next = head;
        while (cur->next && cur->next->next)
        {
            ListNode *move = cur->next->next;
            cur->next->next = move->next;
            move->next = cur->next;
            cur->next = move;
            cur = move->next;
        }
        return dummy.next;
    }
    
    ListNode *swapPairs_2(ListNode *head) {
        if (!head || !head->next) return head;
        ListNode *first = head, *second = head->next;
        first->next = second->next;
        second->next = first;
        first->next = swapPairs(first->next);
        return second;
    }
};