创建动态字符串

Question

我正在尝试构建某种“动态子字符串”，它是由给定string中的循环构建的。规则是我需要按字母顺序找到最长的substring，如果我有一个潮流，我需要计算这两个值，并打印具有较大值的那个。

我读到在python中，字符已经被赋予了一个数值，因此a比b更小；知道这一点，我写了以下内容：

s = "abcsaabcpaosdjaf"
ans = []
# Loop over the string
for i in range(len(s)-1):
    if s[i] < s[i+1]:
        #evaluate if it is in order and build the new string
        ans = s[i]+s[i+1]
    #print the result        
    print(ans)

我的问题是，我不知道如何动态地-我不确定这是否是正确的方式-生成子字符串ans，现在我有了s[i]+s[i+1]，但它只给了我一个包含两个字符的列表，实际上是按字母顺序排列的，并且固定为两个。我怎么做才能让它按原样构建呢？

Answer 1

尝尝这个。希望评论能解释得够多，但如果你不明白，请不要问。

s= "abcsaabcpaosdjaf"
best_answer = ''
current_answer = s[0]
# Loop over the string
for i in s[1:]:
    # look to see if this letter is after the
    # last letter in the current answer.
    if ord(i) > ord(current_answer[-1]):
        # if it is, add the letter to the current
        # answer
        current_answer += i
    else:
        # if it is not, we check if the current
        # answer is longer than the best
        # answer, and update it to the best
        # answer if it is.
        if len(current_answer) > len(best_answer):
            best_answer = current_answer
        # We then set the current answer
        # to just the last letter read.
        current_answer = i

Answer 2

import itertools
s= "abcsaabcpaosdjaf"
result = max(
    (
        list(next(sub)) + [b for a, b in sub]
        for ascending, sub in itertools.groupby(zip(s,s[1:]), lambda x: x[0] <= x[1])
        if ascending
    ),
    key=len
)

print (''.join(result))

致谢给this