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.

Solution

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        dummy = ListNode(0)
        dummy.next = head
        p = dummy
        while p and p.next and p.next.next:
            far_node = p.next.next
            p.next.next = far_node.next
            far_node.next = p.next
            p.next = far_node
            p = p.next.next
        return dummy.next

/**
 * 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) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode current = dummy;

        while(current.next != null && current.next.next != null) {
            ListNode first = current.next;
            ListNode second = current.next.next;

            first.next = second.next;
            second.next = first;
            current.next = second;
            current = current.next.next;
        }

        return dummy.next;
    }
}