python - 如何在列表中按字母顺序取出最长的子串？

Question

比如初始列表L=["by","article","circle","for","line","dance","effort"],我想取出["article","circle","for","line"] 作为结果。因为 L 中的第一个 alpha 是 (b,a,c,f,l,d,e) 并且 (a,c,f,l) 的结果比 (d,e) 长，所以结果是 [ “文章”，“圈子”，“为”，“线”]。如何修复我的代码？我是新手，所以我对 sorted() 和嵌套循环感到很困惑。

def find_longest_consisting(test_case):
    if not test_case:
        return []
    result = []
    for i in range(len(test_case)):
        for j in range(i + len(result) + 1):
            substring = test_case[i:j]
            if len(substring) != (j - i):
                break
            if sorted(substring) == list(substring):
                result = substring
    return result

test_cases = ["by","article","circle","for","line","dance","effort"]
test_result = find_longest_consisting(test_cases)
print(test_result)

Answer 1

最快的方法是维护一个“最长”列表，并在遇到排序变化时将其长度与当前子列表进行比较，如果当前子列表更长，则将其分配给最长的。这只需要对列表进行一次迭代即可获得结果。

def longest_group(lst):
    longest = []
    current = lst[:1]
    for i in range(1, len(lst)):
        if lst[i][0] <= lst[i-1][0]:
            if len(current) > len(longest):
                longest = current
            current = [lst[i]]
        else:
            current.append(lst[i])
    if len(current) > len(longest):
        longest = current
    return longest

print(longest_group(["by","article","circle","for","line","dance","effort"]))
print(longest_group(["zoo","slack","egg"]))

输出:

['article', 'circle', 'for', 'line']
['zoo']

对于您的 ["by","article","circle","for","line","dance","effort"] 列表，这比排序所有的快 10 倍列表中的组合。参见 https://rextester.com/JEJOAE73859用于演示。