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.

一刷
題解：雙指針first和second, 記住要保留first的prev指針信息

public class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode prev = head, first = head, second = head.next;
        //ListNode third = null;
        
        while(second!=null){
            if(first == head) head = second;
            else prev.next = second;
            //third = second.next;
            first.next = second.next;
            second.next = first;
            prev = first;
            
            //
            first = first.next;
            second = first==null? null: first.next;
        }
        return head;
    }
}

二刷：
recursive

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
       if(head == null || head.next == null) return head;
        ListNode n = head.next;
        head.next = swapPairs(head.next.next);
        n.next = head;
        return n;
    }
}