# -*- coding: utf-8 -*-
# @Time    : 2024/4/13 13:23
# @Author  : len
# @File    : 最长公共子串.py
# @Software: PyCharm 
# @Comment :

def longest_common_substring(strs):
    if not strs:
        return ""

    # 定义一个函数来寻找两个字符串的最长公共子串
    def lcs(s1, s2):
        n, m = len(s1), len(s2)
        # 初始化动态规划表
        dp = [[0] * (m + 1) for _ in range(n + 1)]
        longest, x_longest = 0, 0  # 存储最长公共子串的长度和结束位置
        for i in range(1, n + 1):
            for j in range(1, m + 1):
                if s1[i - 1] == s2[j - 1]:  # 当字符相同
                    dp[i][j] = dp[i - 1][j - 1] + 1
                    if dp[i][j] > longest:  # 更新最长公共子串的信息
                        longest = dp[i][j]
                        x_longest = i
                else:
                    dp[i][j] = 0  # 字符不同，公共子串长度重置为0
        return s1[x_longest - longest: x_longest]  # 返回最长公共子串

    # 从列表中的第一个字符串开始，逐步与后续字符串比较
    common_sub = strs[0]
    for s in strs[1:]:
        common_sub = lcs(common_sub, s)
        if not common_sub:  # 如果公共子串长度为0，提前终止
            break

    return common_sub

# 输入的字符串列表
strings = [
    "a1b2c3d4e5",
    "a2b2c3d4e4",
    "a3b3c3d4e6",
    "a5b2c3d4e5"
]

# 调用函数，找出最长的公共子串
result = longest_common_substring(strings)
print(result)