Leetcode - Swap Nodes in Pairs

/**
 * 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) {
        if(head == null)
           return head;
           
        ListNode current = head;
ListNode prev = current.next;
ListNode ln = new ListNode(0);
ListNode tmp = ln;
while (prev != null) {
tmp.next = prev;
tmp = tmp.next;
prev = prev.next;


tmp.next = current;
tmp = tmp.next;
current = prev;


if(prev == null)
tmp.next = null;
else if(prev.next == null){
tmp.next = current;
prev = prev.next;
}
else 
prev = prev.next;
}


return ln.next != null ? ln.next : head;
    }

}

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Hints:

边缘情况，如果末尾是偶数？如果是奇数？如果没有到达末尾。。。。。linkedlist边缘情况够恶心的~

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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.