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num中所有数字能够组成回文数字的总个数。个位数字不被视为回文数字。
【示例】
输入：1221
输出：2
解释：数字1221中，1221整体是回文数字，22也是回文数字，因此总个数为2。
题目难度：中等
题目来源：CodeWars-Numerical Palindrome #3
def solution(num: int)-> int:
# your code here
assert solution(5) == 0
assert solution(1221) == 2
assert solution(141221001) == 5
assert solution(1294) == 0
def solution(num: int)-> int:
if not isinstance(num, int) or num < 0:
return 'Not valid'
nums = str(num)
count = 0
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if nums[i: j + 1] == nums[i: j + 1][::-1]:
count += 1
return count
def solution(num: int)-> int:
# your code here
num_len = len(str(num))
res_list = []
if num_len <= 1:
return 0
for i in range(num_len):
for j in range(i + 2, num_len + 1):
if str(num)[i:j] == str(num)[i:j][::-1]:
res_list.append(str(num)[i:j])
print(res_list)
return len(res_list)
assert solution(5) == 0
assert solution(1221) == 2
assert solution(141221001) == 5
assert solution(1294) == 0
def solution(num: int) -> int:
if len(str(num)) <= 1:
return 0
else:
len_num = len([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]])
if len_num >= 1:
return len_num
else:
return 0
assert solution(5) == 0
assert solution(1221) == 2
assert solution(141221001) == 5
assert solution(1294) == 0
def solution(num: int) -> bool:
num_s = str(num)
num_len = len(num_s)
cnt=0
if num > 10:
for i in range(0, num_len):
min_len = min(i, num_len - i)
for j in range(0, min_len + 1):
if num_s[i - j - 1:i] == num_s[i:i + j + 1][::-1] or num_s[i - j - 1:i] == num_s[i + 1:i + j + 2][::-1]:
cnt+=1
return cnt
def solution2(num: int)-> bool:
num_len = len(str(num))
if num_len <=1:
return 0
else:
s=[True for i in range(num_len) for j in range(i+2,num_len+1) if str(num)[i:j]== str(num)[i:j][::-1] ]
return len(s)
def solution(num: int)->int:
num= str(num)
count = 0
if len(num) < 2:
return count
else:
for i in range(len(num)):
for j in range(i+1,len(num)):
if num[i:1+j] == num[i:1+j][::-1]:
count += 1
return count
def solution_2(num:int)->int:
s = str(num)
n =0
if num >10:
for i in range(len(s)):
for j in range(i+2,len(s)+1):
s1 = s[i:j]
if s1 == s1[::-1]:
n = n+1
return n
def solution(num: int) -> bool:
result = []
length = len(str(num))
for i in range(length):
for j in range(length - i):
ele = str(num)[j:j + i + 1]
if len(ele) <= 1:
pass
# print(f"个位数字{str(num)[j:j + i + 1]}不被视为回文数字。")
elif ele == ele[::-1]:
result.append(ele)
return len(result)
assert solution(5) == 0
assert solution(1221) == 2
assert solution(141221001) == 5
assert solution(1294) == 0
def fun(num: str) -> bool:
if num == ''.join(reversed(num)) and len(num) > 1:
return True
else:
return False
def solution(num: int) -> int:
l = len(str(num))
result = 0
str_num = str(num)
for i in range(l - 1):
for j in range(i + 1, l):
if fun(str_num[i:j + 1]):
result += 1
return result
assert solution(5) == 0
assert solution(1221) == 2
assert solution(141221001) == 5
assert solution(1294) == 0
def solution(num: int) -> int:
# your code here
num_str = str(num)
count = 0
for i in range(len(num_str) - 1):
# print(i)
for j in range(i+1, len(num_str)):
# print(j)
if num_str[i:j+1] == num_str[i:j+1][::-1]:
count += 1
return count
assert solution(5) == 0
assert solution(1221) == 2
assert solution(141221001) == 5
assert solution(1294) == 0
def solution(num: int) -> int:
if len(str(num)) == 1:
return 0
else:
n = 0
for i in range(len(str(num))):
for j in range(i + 1, len(str(num))):
if str(num)[i:j + 1] == str(num)[i:j + 1][::-1]:
n += 1
return n
assert solution(5) == 0
assert solution(1221) == 2
assert solution(141221001) == 5
assert solution(1294) == 0