回文是指一段数字或者文本，正着读和倒着读都是相同的，例如2002，110011都是回文数字。 请编写一个函数，接收一个数字参数num，判断该数字经过重新组合后是否可以变成一个回文数字。 个位数字（0～9）不被视为回文数字。
【示例】
输入：2121
输出：True
解释：数字2121经过重组成1221，它是一个回文数字，因此返回True。
题目难度：中等
题目来源：CodeWars-Numerical Palindrome #5
def solution(num: int)-> bool:
# your code here
assert solution(1331) is True
assert solution(2121) is True
assert solution(3357665) is True
assert solution(13245) is False
solution(num: int)-> bool:
s = str(num)
if not isinstance(num, int) or num < 0:
return "Not valid"
if num > 9 and sum(s.count(x) % 2 for x in set(s)) < 2:
return True
else:
return False
def solution(n):
if n<11:
return False
s=str(n)
times=0
for i in list(set(s)):
If s.count(i)!=0:
times+=1
if times >1:
return False
return True
def solution(num: int)-> bool:
# your code here
if num < 10:
return False
else:
l = []
for i in set(str(num)):
if str(num).count(i) % 2 != 0 and len(str(num)) % 2 == 0:
return False
if str(num).count(i) % 2 != 0 and len(str(num)) % 2 == 1:
l.append(i)
if len(l) > 1:
return False
else:
return True
assert solution(1331) is True
assert solution(2121) is True
assert solution(3357665) is True
assert solution(13245) is False
def solution(num: int) -> bool:
if len(str(num)) <= 1:
return False
else:
list_true = [True for i in range(len(str(num))) for j in range(i + 2, len(str(num)) + 1) if
str(num)[i:j] == str(num)[i:j][::-1] and str(num)[i:j][0] != '0' and str(num)[i:j][
-1] != '0']
if list_true != []:
return True
else:
return False
assert solution(1331) is True
assert solution(2121) is True
assert solution(3357665) is True
assert solution(13245) is False
# Daily_20220412
def solution(num: int)-> bool:
# your code here
if isinstance(num, int) and num > 0:
if num <= 10:
return False
else:
num_list = list(str(num))
repeat_num = set()
for i in range(0, len(num_list)):
if num_list[i] in num_list[i + 1:]:
repeat_num.add(num_list[i])
if len(set(num_list).difference(repeat_num)) <= 1:
return True
else:
return False
else:
return "Not valid"
assert solution(1331) is True
assert solution(2121) is True
assert solution(3357665) is True
assert solution(13245) is False
assert solution(15577666) is False
这种方式还存在bug，晚点重写下。
import itertools
def solution(num: int)-> bool:
if num < 10:
return False
else:
for i in set(''.join(i) for i in itertools.permutations(str(num))):
if i == i[::-1]:
return True
return False
做此题的时，竟感到一丝凄凉
def solution(num: int) -> bool:
if not isinstance(num, int) or num < 10:
return False
# 抓单身🐶啦 T_T，多于一个就报警
count_list = [(str(num).count(n)) % 2 for n in set(str(num))]
return True if count_list.count(1) <= 1 else False
def solution(num: int)-> bool:
# your code here
num = str(num)
a = 0
b = 0
if len(num) == 1:
return False
if int(num) == int(num[::-1]):
return True
for i in set(num):
if num.count(i) % 2 == 0:
a = a + 2
elif num.count(i) % 2 == 1:
b = b + 1
if a != 0 and b == 0:
return True
elif a == 0 and b != 0:
return False
elif a % 2 == 0 and b % 2 ==1:
return True
else:
return False
assert solution(1331) is True
assert solution(2121) is True
assert solution(3357665) is True
assert solution(13245) is False
from itertools import permutations
def fun(num: str) -> bool:
if num == ''.join(reversed(num)) and len(num) > 1:
return True
else:
return False
def solution(num: int) -> bool:
num_list = list(str(num))
perm = permutations(num_list)
for x in perm:
if fun(''.join(x)):
return True
return False
assert solution(1331) is True
assert solution(2121) is True
assert solution(3357665) is True
assert solution(13245) is False
def solution(num: int) -> bool:
num_list = list(str(num))
num_set = set(num_list)
n = 0
for i in num_set:
if num_list.count(i) % 2 != 0:
n += 1
if n > 1:
return False
else:
return True
assert solution(1331) is True
assert solution(2121) is True
assert solution(3357665) is True
assert solution(13245) is False