奇数结点升序偶数结点降序的单链表排序(Python实现)
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题目
一个链表,奇数结点升序,偶数结点降序,要求变成一个全升序的链表。
例如:1->8->2->7->3->6->4->5,变为1->2->3->4->5->6->7->8
解析
按照以下步骤处理:
- 按照奇偶位拆分为两个链表
- 反转偶数结点构成的链表
- 合并两个递增链表
Python实现
# -*- coding:utf-8 -*-
class Node(object):
def __init__(self, val=None, next=None):
self.val = val
self.next = next
def init_list(l):
"""创建不带头结点的单链表"""
head = Node()
tail = head
for val in l:
tail.next = Node(val)
tail = tail.next
tail.next = None
return head.next
def split_list(head):
"""按照奇偶位拆分为两个链表"""
head1 = head2 = None
cur1 = cur2 = None
count = 1
while head:
if count % 2 == 1:
if cur1:
cur1.next = head
cur1 = cur1.next
else:
cur1 = head1 = head
else:
if cur2:
cur2.next = head
cur2 = cur2.next
else:
cur2 = head2 = head
head = head.next
count += 1
cur1.next = None
cur2.next = None
return head1, head2
def reverse_list(head):
"""反转链表"""
if not head or not head.next:
return head
pre = next = None
while head:
next = head.next
head.next = pre
pre = head
head = next
return pre
def merge_list(head1, head2):
"""合并两个递增链表"""
head = Node() # 设置一个临时结点
tail = head
while head1 and head2:
if head1.val <= head2.val:
tail.next = head1
head1 = head1.next
else:
tail.next = head2
head2 = head2.next
tail = tail.next
# 合并剩余结点
if head1:
tail.next = head1
if head2:
tail.next = head2
return head.next
def visit_list(head):
while head:
print(head.val)
head = head.next
if __name__ == ‘__main__‘:
head = init_list([1, 8, 2, 7, 3, 6, 4, 5]) # 创建一个不带头结点的单链表:1->8->2->7->3->6->4->5
head1, head2 = split_list(head) # 1.按照奇偶位拆分为两个链表
head2 = reverse_list(head2) # 2.反转偶数结点构成的链表
head = merge_list(head1, head2) # 3.合并两个递增链表
visit_list(head) # 遍历链表
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