Python单词count

Question

我在学校修了一门Python 2.7课程，他们让我们创建以下程序：

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.

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

Longest substring in alphabetical order is: beggh

In the case of ties, print the first substring.

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

Longest substring in alphabetical order is: abc

我写了以下代码：

s = 'czqriqfsqteavw'

string = ''

tempIndex = 0
prev = ''
curr = ''

index = 0
while index < len(s):
    curr = s[index]
    if index != 0:
        if curr < prev:
            if len(s[tempIndex:index]) > len(string):
               string = s[tempIndex:index]
            tempIndex=index
        elif index == len(s)-1:
            if len(s[tempIndex:index]) > len(string):
               string = s[tempIndex:index+1]
    prev = curr
    index += 1

print 'Longest substring in alphabetical order is: ' + string

老师还给了我们一系列的测试字符串来尝试：

^{pr2}$

除最后一个问题外，所有这些问题都很好，最后一个问题的答案如下：

Longest substring in alphabetical order is: cz

但应该说：

Longest substring in alphabetical order is: avw

我把密码检查了一千遍，没有发现错误。你能帮帮我吗？在

Answer 1

这些线条：

        if len(s[tempIndex:index]) > len(string):
           string = s[tempIndex:index+1]

彼此不同意。如果新的最佳字符串是s[tempIndex:index+1]，那么您应该在If条件中比较的字符串的长度。将它们更改为彼此一致可以解决问题：

^{pr2}$

Answer 2

我看到user5402已经很好地回答了您的问题，但是这个问题引起了我的兴趣，所以我决定重新编写您的代码。：）下面的程序使用与代码基本相同的逻辑，只是做了一些小的更改。在

在实际情况下避免使用索引，并直接在字符串（或其他容器对象）的内容上迭代，被认为更像python。这通常使代码更易于阅读，因为我们不必同时跟踪索引和内容。在

为了访问字符串中的当前和上一个字符，我们压缩输入字符串的两个副本，其中一个副本通过在开头插入空格字符来偏移。我们还将一个空格字符附加到另一个副本的末尾，这样当最长的有序子序列出现在输入字符串的末尾时，我们就不必进行特殊处理。在

#! /usr/bin/env python

''' Find longest ordered substring of a given string 

    From http://stackoverflow.com/q/27937076/4014959
    Written by PM 2Ring 2015.01.14
'''

data = [
    "azcbobobegghakl",
    "abcbcd",
    "onyixlsttpmylw",
    "pdxukpsimdj",
    "yamcrzwwgquqqrpdxmgltap",
    "dkaimdoviquyazmojtex",
    "abcdefghijklmnopqrstuvwxyz",
    "evyeorezmslyn",
    "msbprjtwwnb",
    "laymsbkrprvyuaieitpwpurp",
    "munifxzwieqbhaymkeol",
    "lzasroxnpjqhmpr",
    "evjeewybqpc",
    "vzpdfwbbwxpxsdpfak",
    "zyxwvutsrqponmlkjihgfedcba",
    "vzpdfwbbwxpxsdpfak",
    "jlgpiprth",
    "czqriqfsqteavw",
]


def longest(s):
    ''' Return longest ordered substring of s
        s consists of lower case letters only.
    '''
    found, temp = [], []
    for prev, curr in zip(' ' + s, s + ' '):
        if curr < prev:
            if len(temp) > len(found):
                found = temp[:]
            temp = []
        temp += [curr]
    return ''.join(found)


def main():
    msg = 'Longest substring in alphabetical order is:'
    for s in data:
        print s
        print msg, longest(s)
        print


if __name__ == '__main__':
    main()  

输出

^{pr2}$

Answer 3

指数是你的朋友。 下面是问题的简单代码。在

longword = ''

for x in range(len(s)-1):
    for y in range(len(s)+1):
        word = s[x:y]
        if word == ''.join(sorted(word)):
            if len(word) > len(longword):
                longword = word
print ('Longest substring in alphabetical order is: '+ longword)