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.

Solution1: iterator

Run time complexity is O(n), space complexity is constant space.

/**
 * 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 || head.next == null)
            return head;
        
        ListNode newHead = new ListNode(0);
        newHead.next = head;
        ListNode pre = newHead;
        ListNode cur = head;
        while(cur != null && cur.next != null) {
            ListNode later = cur.next.next;
            ListNode temp = cur.next;
            temp.next = cur;
            pre.next = temp;
            cur.next = later;
            pre = cur;
            cur = later;
        }
        
        return newHead.next;
    }
}

Solution2: recursion

/**
 * 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 null;
        else if(head.next == null)
            return head;
        
        ListNode newHead = head.next;
        ListNode nextPair = head.next.next;
        head.next.next = head;
        head.next = swapPairs(nextPair);
        
        return newHead;
    }
}