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24.py
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52 lines (41 loc) · 1.36 KB
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'''
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.
'''
from ListNode import *
from pprint import pprint
# 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
"""
# 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.
pre = self
pre.next = head # <<<< equals self.next = head
# next attribute is likely not in Solution but setted manually
while pre.next and pre.next.next:
a = pre.next
b = a.next
pre.next, b.next, a.next = b, a, b.next
pre = a
return self.next
def test():
a = [1, 2, 3, 4, 5, 6]
head = List_to_Link(a).head
sl = Solution()
b = sl.swapPairs(head)
Link_to_List(b).print_list()
if __name__ == "__main__":
test()