Leetcode "Swap Nodes in Pairs" Python implementation

title :

Given a linked list, swap every, 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. Modify the values in the list, only nodes itself can be changed.

code : OJ Test via runtime:42 ms

1 #Definition for singly-linked list.2 #class ListNode:3 #def __init__ (self, x):4 #self.val = x5 #Self.next = None6 7 classSolution:8     #@param a ListNode9     #@return a ListNodeTen     defSwappairs (Self, head): One         ifHead isNoneorHead.next isNone: A             returnHead -          -Dummyhead =listnode (0) theDummyhead.next =Head -          -Pre =Dummyhead -Curr =Head +          whileCurr is  notNone andCurr.next is  notNone: -TMP =Curr.next +Curr.next =Tmp.next ATmp.next =Pre.next atPre.next =tmp -Pre =Curr -Curr =Curr.next -         returnDummyhead.next

Ideas :

Basic linked list operations.

It is important to note that while loop judgment condition: First Judge Curr not empty, then judge Curr.next not empty.

This and judgment condition has a short circuit function, and if the Curr is empty it will not make the next judgment, so it is safe

Leetcode "Swap Nodes in Pairs" Python implementation