if string_to_check == string_to_check[::-1]: print("This is a palindrome") else: print("This is not a palindrome" Here, I'm just reversing it, then comparing it with the original Want tolearn pythonin detail? Come and Joinpython courseto gain more knowledge ...
# 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):...
Python Copy 解析:函数名为is_palindrome,参数s为需要检查的字符串,返回值为布尔类型。可以在函数定义处加上类型注释,方便阅读代码。 s[::-1]为字符串切片操作,返回反转后的字符串,通过和原字符串s进行比较,判断是否为回文串。 这个方法非常简洁,但是需要大量的内存来存储反转后的字符串,不适用于处理大型字符串。
function palindromeFn(string) { const stringLength = string.length; for (let i = 0; i < stringLength / 2; i++) { if (string[i] !== string[stringLength - 1 - i]) { return 'It is not a palindrome.'; } } return 'It is a palindrome.'; } console.log(palindromeFn('aviddiva...
A pythonic way to determine if a given value is a palindrome: str(n) == str(n)[::-1] Explanation: We're checking if the string representation of n equals the reversed string representation of n The [::-1] slice takes care of reversing the string After that, we compare for equalit...
boolisPalindrome(stringstr,intlow,inthigh) { // 基本情況 if(low>=high){ returntrue; } return(str[low]==str[high])&& isPalindrome(str,low+1,high-1); } // 檢查是否存在重複子序列的函數 // 在字符串中 boolhasRepeatedSubsequence(stringstr) ...
步骤6:初始化整数n以设置值,并使用if-else语句将参数作为名为isPalimdrome()的调用函数,以检查其是否满足偶数回文数条件,并打印结果。 示例 在下面的示例中,我们将检查偶数位数是否是回文数。 #include<iostream>#include<string>usingnamespacestd;boolisPalindrome(intn){//Using predefined function t...
end()); if (equal(tmp.begin(), tmp.begin() + tmp.size() / 2, tmp.rbegin())) { return true; } else { return false; } } int main() { string s1 = "radar"; string s2 = "Was it a cat I saw"; checkPalindrome(s2) ? cout << "s2 is a palindrome" << endl : cout <<...
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python string 1Answer 0votes answeredFeb 7, 2021byhari_sh(26.4kpoints) The==outweighs in, so you're really trying word==True. >>> w = 'ab c' >>> ' ' in w == True 1: False >>> (' ' in w) == True 2: True However, you don't require==True by any means. in the eve...