没有反向功能的反向字符串[重复]

Posted 2023-02-23

【中文标题】没有反向功能的反向字符串[重复]

【英文标题】：Reverse string without reverse function [duplicate]

【发布时间】：2017-04-10 08:49:10

【问题描述】：

对于输入文本，我必须计算所有元音，将每个单词中的第一个字母大写并反向输出文本（不必大写），而不使用标题或反向函数。我能够算出元音的计数，但与其他两个比较困难。

def main():

    vowelCount = 0
    text = 'abc'

    while(len(text) != 0):
        text = input('Please etner some text or press <ENTER> to exit: ')

        for char in text:
            if char in 'aeiouAEIOU':
                vowelCount += 1
        print('Vowels:', vowelCount)
        vowelCount = 0



        for i in text:
            i = text.find('') + 1
        print(i)

        print(text[0].upper() + text[1:])


main()   

【问题讨论】：

单词倒序还是每个字符倒序？

Reversing a string is already covered in great detail.

【参考方案1】：

这里有两个反转字符串的例子。 切片字符串

>>> s = 'hello'
>>> reversed_s = s[::-1]

或者使用循环。

res = ''
for char in s:
    res = char + res

---

完整代码

def main():
    # Run infinitely until break or return
    # it's more elegant to do a while loop this way with a condition to
    # break instead of setting an initial variable with random value.
    while True:
        text = input('Please enter some text or press <ENTER> to exit: ')
        # if nothing is entered then break
        if not text:
            break

        vowelCount = 0
        res = ''
        prev_letter = None

        for char in text:
            if char.lower() in 'aeiou':
                vowelCount += 1
            # If previous letter was a space or it is the first letter
            # then capitalise it.
            if prev_letter == ' ' or prev_letter is None:
                char = char.upper()

            res += char # add char to result string
            prev_letter = char # update prev_letter

        print(res) # capitalised string
        print(res[::-1]) # reverse the string
        print('Vowel Count is: 0'.format(vowelCount))

# Example
Please enter some text or press <ENTER> to exit: hello world!
Hello World!
!dlroW olleH
Vowel Count is: 3

【讨论】：

感谢提示，但是如何在不反转字符串的情况下使用循环将每个单词大写。

@MrG 更新了答案。我已将其拆分，因此生成的字符串不会反向构造，而是使用建议的第一种方法（切片）在末尾反转