解题思路

  1. 题目要求:

    • 计算两个字符串的最长公共子串长度
    • 子串必须连续
    • 只包含小写字母
    • 字符串长度:
  2. 解题方法:动态规划

    • 使用二维dp数组记录公共子串长度
    • 表示以 结尾的最长公共子串长度
    • 如果当前字符相同,则

代码

def longest_common_substring(str1, str2):
    m, n = len(str1), len(str2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    max_len = 0
    
    # 填充dp数组
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if str1[i-1] == str2[j-1]:
                dp[i][j] = dp[i-1][j-1] + 1
                max_len = max(max_len, dp[i][j])
    
    return max_len

while True:
    try:
        str1 = input().strip()
        str2 = input().strip()
        print(longest_common_substring(str1, str2))
    except:
        break
import java.util.*;

public class Main {
    public static int longestCommonSubstring(String str1, String str2) {
        int m = str1.length(), n = str2.length();
        int[][] dp = new int[m + 1][n + 1];
        int maxLen = 0;
        
        // 填充dp数组
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (str1.charAt(i-1) == str2.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                    maxLen = Math.max(maxLen, dp[i][j]);
                }
            }
        }
        
        return maxLen;
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            String str1 = sc.nextLine();
            String str2 = sc.nextLine();
            System.out.println(longestCommonSubstring(str1, str2));
        }
    }
}
#include <iostream>
#include <string>
#include <vector>
using namespace std;

int longestCommonSubstring(string str1, string str2) {
    int m = str1.length(), n = str2.length();
    vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
    int maxLen = 0;
    
    // 填充dp数组
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (str1[i-1] == str2[j-1]) {
                dp[i][j] = dp[i-1][j-1] + 1;
                maxLen = max(maxLen, dp[i][j]);
            }
        }
    }
    
    return maxLen;
}

int main() {
    string str1, str2;
    while (getline(cin, str1) && getline(cin, str2)) {
        cout << longestCommonSubstring(str1, str2) << endl;
    }
    return 0;
}

算法分析

  • 算法:动态规划
  • 时间复杂度:,其中 是两个字符串的长度
  • 空间复杂度:,需要二维dp数组