Swap Nodes in Pairs

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.

Really a hard one for me, because I am not sure the pointers direction...

1. new ListNode p, p.next = head;

2. two pointers, curr= head; prev = p;

3. when loop begin, two pointers to track pair: first and second.

Failed with recursion way...

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {/*    public ListNode swapPairs(ListNode head) {        ListNode p = new ListNode(0);        p.next = head;        swapParis(p);           return p.next;    }    public ListNode swapParis(ListNode n){         if(n == null || n.next == null || n.next.next==null){            return n;        }        ListNode tmp = n.next.next;        ListNode ptmp = n.next;        n.next.next = n;        n.next = swapPairs(tmp);        return ptmp;    }*/public ListNode swapPairs(ListNode head){    ListNode p = new ListNode(0);    p.next = head;    ListNode prev = p;    ListNode curr = head;       if(head==null || head.next == null){        return p.next;    }    while(curr!=null && curr.next!=null){        ListNode first = curr;        ListNode second = curr.next;        curr = curr.next.next;        prev.next = second;        second.next = first;        first.next = curr;        prev = first;    }    return p.next;}}