Python：如何查找字符串中字母序连续子串并统计其长度？

需求 #

编写Python程序，提取输入字符串中按字母顺序连续排列的子串，并统计每个子串的字符数量：

• 输入cabin，输出abc, 3
• 输入sightfulness，输出ghi, 3、stu, 3

当前进度与问题 #

已完成输入字符串转字符列表、去重、排序操作，但无法正确检查排序后列表中字母是否连续，当前代码如下：

import string

a = input("Input A: ")

# sorted_a is the sorted letters of the string input a
sorted_a = sorted(a)
print(sorted_a)

# to remove the duplicate letters in sorted_a
# make a temporary list to contain the filtered elements
temp = []
for x in sorted_a:
    if x not in temp:
        temp.append(x)

# pass the temp list to sorted_a, sorted_a list updated
sorted_a = temp
joined_a = "".join(sorted_a)
print(sorted_a)


alphabet = list(string.ascii_letters)
print(alphabet)

def check_list_order(sorted_a):
    in_order_list = []
    for i in sorted_a:
        if any((match := substring) in i for substring in alphabet):
            print(match)

            # this should be the part
            # that i would compare the element
            # in sorted_a with the elements in alphabet
            # to know the order of both of them
            # and to put them ordered characters
            # to in_order_list
            
            if ord(i)+1 == ord(i)+1:
                in_order_list.append(i)
    return in_order_list

print(check_list_order(sorted_a))

修正方案 #

原代码核心问题是连续字母的判断逻辑错误，可通过直接比较字符的ASCII码值判断连续性。同时简化去重排序步骤，以下是修正后的完整代码：

def find_consecutive_substrings(s):
    # 一步完成去重+按字母顺序排序
    sorted_chars = sorted(set(s))
    if not sorted_chars:
        return []
    
    result = []
    current_sub = [sorted_chars[0]]
    
    for char in sorted_chars[1:]:
        # 判断当前字符是否是前一个字符的下一个字母
        if ord(char) == ord(current_sub[-1]) + 1:
            current_sub.append(char)
        else:
            # 仅收集长度≥2的连续子串（匹配示例需求）
            if len(current_sub) >= 2:
                result.append(("".join(current_sub), len(current_sub)))
            current_sub = [char]
    # 处理循环结束后剩余的连续子串
    if len(current_sub) >= 2:
        result.append(("".join(current_sub), len(current_sub)))
    return result

# 执行逻辑
input_str = input("Input A: ")
substrings = find_consecutive_substrings(input_str)
for sub, length in substrings:
    print(f"{sub}, {length}")

关键逻辑说明 #

1. 去重排序优化：用sorted(set(s))替代原循环去重，代码更简洁高效
2. 连续性判断：通过ord(char) == ord(current_sub[-1]) + 1直接校验字符是否连续
3. 子串收集：用临时列表跟踪当前连续序列，遇到断点时保存有效子串，最后处理剩余序列
4. 输出适配：按示例格式输出每个子串和对应长度

内容的提问来源于stack exchange，提问作者Kite