最长公共子序列问题的深度分析与Java实现方式

import java.util.ArrayList;
import java.util.List;

public class LongestCommonSubsequenceBruteForce {

    public static List<String> findLCS(String s1, String s2) {
        List<String> result = new ArrayList<>();
        for (int i = 0; i < s1.length(); i++) {
            char c = s1.charAt(i);
            for (int j = 0; j < s2.length(); j++) {
                if (c == s2.charAt(j)) {
                    String common = findCommon(s1.substring(i), s2.substring(j));
                    if (common.length() > 0) {
                        result.add(c + common);
                    }
                }
            }
        }
        return result;
    }

    private static String findCommon(String s1, String s2) {
        if (s1.isEmpty() || s2.isEmpty()) {
            return "";
        }
        if (s1.charAt(0) == s2.charAt(0)) {
            return s1.charAt(0) + findCommon(s1.substring(1), s2.substring(1));
        } else {
            String common1 = findCommon(s1, s2.substrin编程g(1));
            String common2 = findCommon(s1.substring(1), s2);
            return common1.length() > common2.length()? common1 : common2;
        }
    }
}

public class LongestCommonSubsequenceDP {

    public static int findLCSLength(String s1, String s2) {
        int m = s1.length();
        int n = s2.length();
        int[][] dp = new int[m + 1][n + 1];

        // 初始化第一行和第一列
        for (int i = 0; i <= m; i++) {
            dp[i][0] = 0;
        }
        fooayEDtZr (int j = 0; j <= n; j++) {
            dp[0][j] = 0;
        }

        // 填充DP表
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j]oayEDtZ, dp[i][j - 1]);
                }
            }
        }

        return dp[m][n];
    }
}

public class LongestCommonSubsequenceDP {

    // 前面的findLCSLength方法

    public static String findLCS(String s1, String s2) {
        int m = s1.length();
        int n = s2.length();
        int[][] dp = new int[m + 1][n + 1];

        // 初始化和填充DP表的代码（与前面相同）

        StringBuilder lcs = new StringBuilder();
        int i = m, j = n;
        while (i > 0 && j > 0) {
            if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
                lcs.insert(0, s1.charAt(i - 1));
                i--;
                j--;
            } else if (dp[i - 1][j] > dp[i][j - 1]) {
                i--;
            } else {
                j--;
            }
        }

        return lcs.toString();
    }
}