Question

我试图构建某种＆＃34;动态子字符串＆＃34;这是在给定string的循环中构建的。规则是我需要按字母顺序找到最长的substring，如果我有潮流，我需要对两者进行评估并打印出价值较大的那个。

我读过在python中已经给出了数字值，因此a低于b;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)

我遇到的问题是我不知道如何动态 - 我不确定这是否是正确的说法 - 构建子串ans，现在我有{{ 1}}但是这只给了我一个实际上按字母顺序排列的两个字符的列表，它只固定为两个。我怎么能以它构建它的方式来做它？

Answer 1

试试这个。希望能够充分解释这些评论，但如果您不理解，请随便提出。

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this

的信用

创建一个动态字符串

2 个答案: