编写打印字母顺序最长的字符串的代码

Question

我正在尝试编写符合标题的代码。如果两个字符串是相同的大小，只需打印出第一串

例如：

s = 'abcbcd' 

会打印出

'abc' 

这是我到目前为止有：

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：

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：为了详细说明为什么你的代码没有产生正确的结果 - 你与字母try语句追加订单不分胜负。因此，如果try语句返回true，则字母会附加两次，如果不是，则字母仍会附加。 编辑：固定码

Answer 2

尝试用这种（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) 

看timgeb在我连着一个完整的答案的链接的答案。

Answer 3

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