# Python program to check prime number # Function to check prime number def isPrime(n): return all([(n % j) for j in range(2, int(n/2)+1)]) and n>1 # Main code num = 59 if isPrime(num): print(num, "is a prime number") else: print(num, "is not a prime number") ...
Python program to check prime number using object oriented approach# Define a class for Checking prime number class Check : # Constructor def __init__(self,number) : self.num = number # define a method for checking number is prime or not def isPrime(self) : for i in range(2, int(...
# Program to check if a string is palindrome or not my_str = 'aIbohPhoBiA' # make it suitable for caseless comparison my_str = my_str.casefold() # reverse the string rev_str = reversed(my_str) # check if the string is equal to its reverse if list(my_str) == list(rev_str):...
class Solution(object): def isPalindrome(self, x): """ :type x: int :rtype: bool """ if x < 0: return False temp = x y = 0 while temp: y = y*10 + temp%10 temp /= 10 return x == y 总结 由于不允许占用额外空间,所以不能将其分为字符串来做。 本文参与 腾讯云自媒体同步曝...
Here is source code of the Python Program to check whether a string is a palindrome or not using recursion. The program output is also shown below. def is_palindrome(s): if len(s) < 1: return True else: if s[0] == s[-1]: return is_palindrome(s[1:-1]) else: return False ...
Check Palindrome in Python Using List Slicing Example # Enter stringword=input()# Check for palindrome strings using list slicingifstr(word)==str(word)[::-1]:print("Palindrome")else:print("Not Palindrome") The program begins by prompting the user to input a string using theinput()function...
Determine whether an integer is a palindrome. Do this without extra space. 这一题描述很简单,判断一个数字是否是回文数,不要用额外的空间。其实我觉得逻辑不是那么简答,2个点需要注意,怎样的数字是回文数和怎样算不使用额外空间? 1、先考虑什么数字是回文数?
Program to add two numbers in Python <br> a = input('Enter first number: ')<br> b = input('Enter second number: ')<br> sum = float(a) + float(b)<br> print('The sum of {0} and {1} is {2}'.format(num1, num2, sum))<br> Program to Check Armstrong Number in Python...
一次AC题目要求中有空间限制,因此没有采用字符串由量变向中间逐个对比的方法,而是采用计算翻转之后的数字与x是否相等的方法; 1 class Solution: 2 # @return a boolean 3 def isPalindrome(self, x): 4 o...
所谓回文数 Palindrome Number,即从左边开始读或从右边开始读,两者结果一致。判断的目标数字为整数,包括负数。 比如12321,123321,或者 3,都是回文数。 -12321不是回文数;-1也不是回文数。 解法1. 简单解法:将整数转换为字符串 转换之后,Python有转换的 reverse 函数,将字符串进行反转:str[::-1]。