How to compare two linked lists are equal in Python All In One
在 Python 中如何比较两个链表是否相等
# Definition for singly-linked list.
from typing import Optional
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
# DFS / BFS
def getLength(node):
temp = node
length = 1;
while temp.next != None:
length += 1
temp = temp.next
return length
def dfs(node, length):
temp = node
# 向下取整除法
l = length // 2
while l > 0:
l -= 1
temp = temp.next
return temp
return dfs(head, getLength(head))
def SingleLinkedListGenerator(num: int) -> Optional[ListNode]:
head = ListNode()
temp = head
for i in range(num):
temp.next = ListNode(i + 1)
temp = temp.next
return head.next
i = SingleLinkedListGenerator(7)
r = [4,5,6,7]
print("input ", i)
# how to compare two linked lists are equal in Python ??? JSON.stringify(list)
"""
$ py3 ./876_middle-of-the-linked-list.py
"""
https://leetcode.com/problems/middle-of-the-linked-list/
# # test cases
# t1 = 14
# t2 = 8
# t3 = 123
# tests = [t1, t2, t3]
# r1 = 6
# r2= 4
# r3 = 12
# results = [r1, r2, r3]
# def test():
# solution = Solution()
# for index, test in enumerate(tests):
# value = solution.numberOfSteps(test)
# result = results[index]
# if(str(value) == str(result)):
# print("✅ passed", index)
# else:
# print("❌ failed", value, result)
# test()
solutions
def equal(self, other):
while self and other and self.value == other.value:
self = self.next
other = other.next
if not self and not other:
return True
return False
def check_equality(list_1, list_2):
curr_1 = list_1.head
curr_2 = list_2.head
while (curr_1 and curr_2):
if curr_1.data != curr_2.data:
return False
curr_1 = curr_1.next
curr_2 = curr_2.next
if curr_1 is None and curr_2 is None:
return True
else:
return False
json
import json
list1 = [1,2,3]
list2 = [1,2,3]
result = json.dumps(list1) == json.dumps(list2)
print("result =", result)
"""
$ py3 ./json.stringify_json.dumps.py
result = True
"""
https://www.cnblogs.com/xgqfrms/p/17626511.html
demos
# Definition for singly-linked list.
from typing import Optional
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
# DFS / BFS
def getLength(node):
temp = node
length = 1;
while temp.next != None:
length += 1
temp = temp.next
return length
def dfs(node, length):
temp = node
# 向下取整除法
l = length // 2
while l > 0:
l -= 1
temp = temp.next
return temp
return dfs(head, getLength(head))
# def SingleLinkedListGenerator(num: int) -> Optional[ListNode]:
# head = ListNode()
# temp = head
# for i in range(num):
# temp.next = ListNode(i + 1)
# temp = temp.next
# return head.next
def SingleLinkedListGenerator(num: int, start: int = 1) -> Optional[ListNode]:
head = ListNode()
temp = head
for i in range(num):
if(start == 1):
temp.next = ListNode(i + start)
# print("temp.next.val =", temp.next.val)
temp = temp.next
else:
if(i < start):
temp.next = ListNode(i + start)
# print("temp.next.val =", temp.next.val)
temp = temp.next
return head.next
def compareListEqual(one, two):
while one and two and one.val == two.val:
one = one.next
two = two.next
# 同时为 None ✅
if not one and not two:
return True
return False
input = SingleLinkedListGenerator(7)
# print("\n")
result = SingleLinkedListGenerator(7, 4)
# i = [1,2,3,4,5,6,7]
# r = [4,5,6,7]
solution = Solution()
case = solution.middleNode(input)
test = compareListEqual(case, result)
if test:
print("✅ test =", test)
else:
print("❌ test =", test)
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refs
https://stackoverflow.com/questions/54388326/how-do-i-compare-two-linked-lists-in-python-3-7
https://www.tutorialspoint.com/python-program-to-check-whether-2-linked-lists-are-same
https://www.geeksforgeeks.org/python-program-to-check-if-two-linked-lists-are-identical/
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