DAY12：leetcode #24 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.

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# 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
"""
if head == None:
return None
if head.next == None:
return head
temp = head.next
head.next = head.next.next
temp.next = head
head = temp
pointer_list = [head,head.next,head.next.next]
if pointer_list[2]:
pointer_list.append(pointer_list[2].next)
del pointer_list[0]
else:
return head
while pointer_list[2]:
temp = pointer_list[0].next
temp.next = pointer_list[2].next
pointer_list[0].next = pointer_list[2]
pointer_list[2].next = temp
pointer_list.append(pointer_list[1].next)
del pointer_list[0]
if pointer_list[2]:
pointer_list.append(pointer_list[2].next)
del pointer_list[1]
else:
break
return head


代码比较啰嗦，有机会简化一下