最长公共子串是指在两个字符串中同时出现的最长连续子字符串。与最长公共子序列(Longest Common Subsequence)不同,子串要求是连续的,而子序列可以是不连续的。举个例子,给定字符串 “abcdxyz” 和“xyzabcd”。它们的最长公共子串是 “abcd”。 算法思路 常见的解决该问题的算法有动态规划(Dynamic Programming)。我...
//边界 for(inti=0;i<=lenA;++i) { dp[i][0]=0; } for(intj=0;j<=lenB;++j) { dp[0][j]=0; } //状态转移方程 for(inti=1;i<=lenA;++i) { for(intj=1;j<=lenB;++j) { if(A[i]==B[j]){ dp[i][j]=dp[i-1][j-1]+1; }else{ dp[i][j]=max(dp[i-1][j],dp...
class Solution: def longestCommonSubsequence(self, text1: str, text2: str) -> int: m, n = len(text1), len(text2) # dp[i][j] 表示 text1[..i] 和 text2[..j] 的最长公共子序列的长度 dp = [[0] * (n + 1) for _ in range(m + 1)] for i in range(m): for j in ra...
This method is a recursive approach to solving the LCS problem in Python. It checks for all possible subsequences of given sequences and finds the longest common subsequence. Dynamic programming is the optimization of the plain recursion method. As we can see, there are overlapping sub-problems ...
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Exercise 9: Python Code for the Longest Subsequence Length Given an unsorted array of integers, find the length of the longest increasing subsequence. In an array, a “Longest Increasing Subsequence” is a sequence where each number is greater than the previous one. They don’t have to be ne...
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def longest_common_prefix(strs): if not strs: return "" prefix = strs[0] for i in range(1, len(strs)): while strs[i].find(prefix) != 0: prefix = prefix[:-1] if not prefix: return "" return prefixprint(longest_common_prefix(["flower", "flow", "flight"])) # 输出: "...
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