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Copy path143_reorder_list.py
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143_reorder_list.py
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# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reorderList(self, head: ListNode) -> None:
"""
Do not return anything, modify head in-place instead.
"""
def find_mid_node(head: ListNode) -> ListNode:
slow = fast = head
while (fast != None) and (fast.next != None):
slow = slow.next
fast = fast.next.next
return slow
def reverse_list(head: ListNode) -> ListNode:
if (head == None) or (head.next == None):
return head
curr_pre = None
curr = head
curr_next = curr
while curr != None:
curr_next = curr.next
curr.next = curr_pre
curr_pre = curr
curr = curr_next
return curr_pre
if (head == None) or (head.next == None):
return
mid_node = find_mid_node(head)
second_half = mid_node.next
second_half = reverse_list(second_half)
mid_node.next = None
p1 = head
p2 = second_half
while (p1 != None) and (p2 != None):
tmp1 = p1.next
tmp2 = p2.next
p1.next = p2
p1 = tmp1
p2.next = p1
p2 = tmp2
'''
This is my personal record of solving Leetcode Problems.
If you have any questions, please discuss them in [Issues](https://github.com/mengxinayan/leetcode/issues).
Copyright (C) 2020-2022 mengxinayan
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation, either version 3 of the License, or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program. If not, see <http://www.gnu.org/licenses/>.
'''