string = ''.join(e for e in string if e.isalnum())将字符串转换为小写 string = string.lower()判断字符串是否为回文串 return string == string[::-1]测试 print(is_palindrome("A man, a plan, a canal, Panama!")) # True print(is_palindrome("race a car")) # False 程...
Check if a string is palindrome ignoring whitespace and special characters 6 Determining if string is palindrome 3 Checking Palindrome text, with ignored punctuation marks, spaces and case 6 How to check if a string is a palindrome? 0 To check whether palindrome or not 3...
# 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):...
if lenStr < 2:return True elif string[0] != string[-1]:return False else:return is_string_palindrome(string[1:-1])# 获取用户输入 user_input = input()# 调用函数 print(is_string_palindrome(user_input.lower()))3、代码分析:该题可以用递归的方法实现,不断的去掉头尾后继续判断,递归结束条...
if letter=='e': return False return True has_no_e('tdy') 1. 2. 3. 4. 5. 6. 但是总是跟预想的结果不一样,单词有‘e'时也返回True,所以添加一个print想看看情况: def has_no_e(word): for letter in word: if letter=='e':
my_string = "abcba"if my_string == my_string[::-1]:print("palindrome")else: print("not palindrome")# Output# palindrome 1. 统计列表元素的个数 有多种方式可以实现这个技巧,但我最喜欢的是采用 Counter 类。 Counter 可以统计给定列表中每个元素的个数,返回一个字典格式。示例如下,其中 most_common...
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...
证明:0+1+2+3+……+n = n(n+1)/2 Proof: If n=0, then LHS is 0 and RHS is 0*1/2=0, so true Assume true for some k, then need to show that 0+1+2+...+k+(K+1) = (k+1)(k+2)/2 LHS is k(k+1)/2+k+1 = RHs ...
2116 Check if a Parentheses String Can Be Valid C++ Python O(n) O(1) Medium 2124 Check if All A's Appears Before All B's C++ Python O(n) O(1) Easy 2129 Capitalize the Title C++ Python O(n) O(1) Easy 2131 Longest Palindrome by Concatenating Two Letter Words C++ Python O(...
Check if a string follows a n b n pattern or not in Python - Suppose we have a string s we have to check whether the string is following the pattern a^nb^n or not. This is actually a string when n = 3, the string will be aaabbb.So, if the input is like s