Python code to check if a number is a palindrome using iteration: def is_palindrome(num): str_num = str(num) start = 0 end = len(str_num) - 1 while start < end: if str_num[start] != str_num[end]: return False start += 1 end -= 1 return True print(is_palindrome(121)) ...
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]。 代码如下: ...
# Enter input number or stringword=input()# Check if the string is a palindrome, ignoring case and spacesdefis_palindrome(s):s="".join(c.lower()forcinsifc.isalnum())returns==s[::-1]# Check and print the resultifis_palindrome(word):print("Palindrome")else:print("Not Palindrome") ...
leetcode:Palindrome Number【Python版】 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; 1classSolution:2#@return a boolean3defisPalindrome(self, x):4o =x5ret =06flag = 17ifx <0:8returnFalse9while(x!=0):10ret = ret*...
12 13 14 15 16 17 18 19 20 21 classSolution(object): defisPalindrome(self, x): ifx<0:returnFalse i,a=0,x whilea!=0: a=a/10 i+=1 first_i=10**(i-1) last_i=10 whilefirst_i>=last_i: first=x/first_i last=x%last_i ...
Palindrome Number (回文数) 题目Determine whether an integer is a palindrome. Do this without extra space. 什么是回文数 来源于百度百科:回文数 “回文”是指正读反读都能读通的句子,它是古今中外都有的一种修辞方式和文字游戏,如“我为人人,人人为我”等。在数学中也有这样一类数字有这样的特征,成为...
Given a number n, size of list then next line contains space separated n numbers.LogicWe will simply convert the number into string and then using reversed(string) predefined function in python ,we will check whether the reversed string is same as the number or not....
Check Whether a String is Palindrome or Not Remove Punctuations From a String Sort Words in Alphabetic Order Illustrate Different Set Operations Count the Number of Each Vowel Python Tutorials Python List reverse() Python reversed() Python String casefold() Python List sort() Python ...
# Python program to check if a string is # palindrome or not # function to check palindrome string def isPalindrome(string): rev_string = string[::-1] return string == rev_string # Main code x = "Google" if isPalindrome(x): print(x,"is a palindrome string") else: print(x,"is...
python3 class Solution: """ @param num: a positive number @return: true if it's a palindrome or false """ def isPalindrome(self, num): # write your code herelist_num=list(str(num)) new_list_num=[]foriinrange(len(list_num)):iflist_num[i]==list_num[len(list_num)-i-1]: ...