For future visitors, these don't address all the constraints in the problem, but the easiest way to check if a string is a palindrome in python is to simply do: defispal(s):returns == s[::-1] string):returnall(x == yforx, yinzip(string,reversed(string))) Thezip()function itera...
57 How to check for palindrome using Python logic 0 Python 2.7.6: Making a palindrome checker with a for-loop 0 return result of palindrome check with for-loop in python 0 How to check for palindrome using while loop python 0 I need help finishing my palindrome checking ...
Check Palindrome String in Python Using a Loop Example # Enter string word = input() # Check if string is palindrome using a loop is_palindrome = True length = len(word) for i in range(length // 2): if word[i] != word[length - i - 1]: is_palindrome = False break if is_pali...
使用forloop的scala回文函数 、、 我想知道Scala中Java回文函数的等价物,在scala中编写带有多个变量的forloop是很棘手的 class Solution { public boolean isPalindrome(String s) { for (int i = 0, j = s.length() - 1; i < j; i++, j--) { while (i < j && !Character.isLetterOrDigit(s....
Given an integer number and we have to check whether it is palindrome or not using java program.Check palindrome number in javaimport java.util.Scanner; class CheckNumberPalindromeOrNotPalindromeClass{ public static void main(String[] args){ //rev variable is used to store reverse of actual_...
If there is no break statement after the loop has finished, the number is a palindrome. 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...
one line in pythoninput == input[::-1]ReplyMartijn.sMay 6, 2019 at 11:51 AM please don't teach anyone to reverse strings using recursion, or worse, check if they are a palindrome by reversing and comparing the string, if an interviewer asks me to do this i will tell him its a...
In this section, we will learn how to reverse a number in Java using while loop, for loop and recursion. To reverse a number, follow the steps given below: First, we find the remainder of the given number by using the modulo (%) operator. Multiply the variable reverse by 10 and add...
A simple solution in Swift: func isPalindrome(word: String) -> Bool { // If no string found, return false if word.count == 0 { return false } var index = 0 var characters = Array(word) // make array of characters while index < characters.count / 2 { // repeat l...
A naive solution in Python would be to useitertools.permutations(words, len(words))to check every possible solution, but as word count grows, I believe this is at leastO(n!*c)with n words and c characters. Heap's algorithm doesn't seem to particularly lend itself we...