• 分類：LinkedList

• 考察知識(shí)點(diǎn)：LinkedList

• 最優(yōu)解時(shí)間復(fù)雜度：**O(n) **

24. Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

Note:

• Your algorithm should use only constant extra space.
• You may not modify the values in the list's nodes, only nodes itself may be changed.

代碼：

我的解法:

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

class Solution:
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head==None or head.next==None:
            return head
        
        dummy=ListNode(0)
        dummy.next=head
        l1=dummy
        l2=head
        
        while(l2!=None and l2.next!=None):
            
            new_head=l2.next.next
            
            l1.next=l2.next
            l2.next.next=l2
            l2.next=new_head
            
            l1=l2
            l2=l2.next
            
            
        return dummy.next

討論：

1.這道題在面試中出現(xiàn)頻率挺低的
2.這道題主要就是為了讓人搞明白鏈表的next到底是啥，會(huì)不會(huì)被繞暈（事實(shí)上我就覺得非常的暈，暈上天了@.@）

過程圖

Outcome