Question

我正在尝试编写标题所说的代码。如果两个子串的大小相同，只需打印第一个字符串

例如：

s = 'abcbcd'

会打印出来

'abc'

这是我到目前为止所做的：

old_order = ''
re = ''

for r in range(len(s)):
    order = ''
    for letter in s[r:]:
        try:
            if letter <= s[s.index(letter)+1]:
                order += letter 
        except:
            order += letter
        else:
            order += letter
            print(order)
            if r is 0:
                old_order = order
            if len(order) > len(old_order):
                re = order

            else:
                re = old_order
print(re)

我得到了什么：

aabbcbbc

Answer 1

def isInAlphabeticalOrder(word):
    preString=''
    alphaString=''
    for char in word:
        if(alphaString==''):
            alphaString=char
        elif(alphaString[-1]<=char):
            alphaString+=char
        elif(alphaString[-1]>char):
           if(len(preString)<len(alphaString)):
               preString=alphaString
               alphaString=char
           else:
               alphaString=char
    if(len(alphaString)>len(preString)):
        preString=alphaString
    print(preString)

编辑：详细说明为什么您的代码没有产生正确的结果 - 无论try语句中的结果如何，您都在附上该字母的顺序。因此，如果try语句返回true，则会附加两次字母，否则，附加字母。编辑：固定代码

Answer 2

代码：

def longest_sub_str(A):
    final_res = ""
    tmp = ""
    prev = ""
    for i in A:
        if prev < i:
            prev = i
            tmp += i
        else:
            prev = ""
            if len(tmp)>len(final_res):
                final_res = tmp
            tmp = ""
    return tmp if len(tmp)>len(final_res) else final_res


print longest_sub_str('abcdddabcddefbcd')

输出：

abcd

Answer 3

试试这个（finding the longest substring of letters in alphabetical order in a given string）：

s = 'abcbcd'
r = []
substring = []
for length in s:
    r = r + [length]        
    if r == sorted(r) and len(r) >= len(substring):
        substring=r
    elif r != sorted(r):
        r = [r[len(r)-1]]        

print(substring)

在我附上的链接中查看 timgeb 的答案以获得完整答案。

编写打印字符串字符顺序的字符串最长子字符串的代码

3 个答案: