48 lines
1.5 KiB
Python
48 lines
1.5 KiB
Python
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# -*- coding: utf-8 -*-
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# @Time : 2024/4/13 13:23
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# @Author : len
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# @File : 最长公共子串.py
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# @Software: PyCharm
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# @Comment :
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def longest_common_substring(strs):
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if not strs:
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return ""
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# 定义一个函数来寻找两个字符串的最长公共子串
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def lcs(s1, s2):
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n, m = len(s1), len(s2)
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# 初始化动态规划表
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dp = [[0] * (m + 1) for _ in range(n + 1)]
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longest, x_longest = 0, 0 # 存储最长公共子串的长度和结束位置
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for i in range(1, n + 1):
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for j in range(1, m + 1):
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if s1[i - 1] == s2[j - 1]: # 当字符相同
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dp[i][j] = dp[i - 1][j - 1] + 1
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if dp[i][j] > longest: # 更新最长公共子串的信息
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longest = dp[i][j]
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x_longest = i
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else:
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dp[i][j] = 0 # 字符不同，公共子串长度重置为0
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return s1[x_longest - longest: x_longest] # 返回最长公共子串
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# 从列表中的第一个字符串开始，逐步与后续字符串比较
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common_sub = strs[0]
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for s in strs[1:]:
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common_sub = lcs(common_sub, s)
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if not common_sub: # 如果公共子串长度为0，提前终止
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break
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return common_sub
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# 输入的字符串列表
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strings = [
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"a1b2c3d4e5",
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"a2b2c3d4e4",
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"a3b3c3d4e6",
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"a5b2c3d4e5"
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]
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# 调用函数，找出最长的公共子串
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result = longest_common_substring(strings)
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print(result)
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