Swap Nodes in Pairs Leetcode Python

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.

与大部分linklist的题目一样，这道题也要定义一个dummy head 将dummy head 设为新的头

一共需需要两个辅助指针。时间复杂度为O(n)

先将p指向 dummy head 然后当p.next 以及p.next.next存在的时候将两两进行调换

代码如下

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    # @param a ListNode    # @return a ListNode    def swapPairs(self, head):        if head==None or head.next==None:            return head        dummy=ListNode(0)        dummy.next=head        p=dummy        while p.next and p.next.next:            tmp=p.next.next            p.next.next=tmp.next            tmp.next=p.next            p.next=tmp            p=p.next.next        return dummy.next