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.

To go from pre -> a -> b -> b.next to pre -> b -> a -> b.next, we need to change those three references. Instead of thinking about in what order I change them, I just change all three at once.

# 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
        """
        pre = res = ListNode(0)
        pre.next = head
        res.next = head
        while pre.next and pre.next.next:
            a = pre.next
            b = pre.next.next
            pre.next, b.next, a.next = b, a, b.next
            pre = a
        return res.next