使用 Python 打印最长的字母子字符串，对于并列，打印第一个子字符串 [英] Print the longest alphabetical substring using Python and for ties, print the first substring

问题描述

假设 s 是一串小写字符.编写一个程序，打印 s 中字母按字母顺序出现的最长子串.

Assume s is a string of lower case characters. Write a program that prints the longest substring of s in which the letters occur in alphabetical order.

例如，如果 s = 'azcbobobegghakl'，那么你的程序应该打印

For example, if s = 'azcbobobegghakl', then your program should print

按字母顺序排列的最长子串是:beggh

Longest substring in alphabetical order is: beggh

在平局的情况下，打印第一个子字符串.例如，如果 s = 'abcbcd'，那么你的程序应该打印

In the case of ties, print the first substring. For example, if s = 'abcbcd', then your program should print

按字母顺序排列的最长子串是:abc

Longest substring in alphabetical order is: abc

这是我找到的代码.我如何在上面给出的关于关系的问题中实现后一个条件?

Here's the code I found. How do I implement the latter condition in the question given above regarding ties?

    *s = raw_input('provide string: ')
    result = []
    final = []
    for letters in s:
        result = result + [letters]        
        if result == sorted(result) and len(result) >= len(final):
            final = result            
        elif result != sorted(result):
            result = [result[len(result)-1]]        
    print('Longest substring in alphabetical order is: '+(''.join(final)))*

推荐答案

我会通过以下方式解决问题:

I'd approach the problem the following way:

• 让我们定义两个字符串:递增字母的 current 字符串和当前 longest 字符串.
• 两个字符串都用第一个字母初始化.(这样我们就可以随时阅读他们的最后一封信.)
• 然后我们遍历输入字符串s(从第二个字符开始).
• 如果当前字符 c 满足要求 c >= current[-1]，我们将其添加到当前解决方案中.
• 我们可能将 current 字符串存储为 longest.
• 如果 c 不满足排序要求，我们从一个新的解决方案 current = c 开始.
• 最后，我们打印longest 字符串.

• Let's define two strings: The current string of increasing letters and the currently longest string.
• Both strings are initialized with the first letter. (This way we can always read their last letter.)
• Then we iterate over the input string s (starting with the second character).
• If the current character c fulfills the requirement c >= current[-1], we add it to the current solution.
• We possibly store the current string as longest.
• If c does not fulfill the ordering requirement, we start with a new solution current = c.
• Finally, we print the longest string.

s = "azcbobobegghakl"
longest = s[0]
current = s[0]
for c in s[1:]:
    if c >= current[-1]:
        current += c
        if len(current) > len(longest):
            longest = current
    else:
        current = c
print "Longest substring in alphabetical order is:", longest

<小时>

如何修复您的代码.提到的条件:

在比较len(result) >= len(final)时使用>代替>=，即只更新final 解决方案，如果它更长，但不是如果它具有相同的长度.

Use > instead of >= in the comparison len(result) >= len(final), i.e. only update the final solution if it is longer, but not if it has the same length.

考虑 Dylans 的评论

你说得对.我更新了代码和描述以正确处理 s 以最长的字母子字符串结尾的情况.(将 else: 向下移动两行就足够了.)

You're right. I updated both code and description to correctly handle the case when s ends with the longest alphabetical substring. (Moving else: two lines down was sufficient.)

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