Question

我们可以通过DP（动态编程）找到两个字符串的LCS（最长公共子序列）。通过跟踪DP表，我们可以获得LCS。但是，如果存在多个LCS，我们怎么能得到它们呢？

示例：

string1 : bcab
string2 : abc

这里“ab”和“bc”都是LCS。

Answer 1

这是一个有效的java解决方案。有关解释，您可以看到我的答案 How to print all possible solutions for Longest Common subsequence

static int arr[][];
static void lcs(String s1, String s2) {
    for (int i = 1; i <= s1.length(); i++) {
        for (int j = 1; j <= s2.length(); j++) {
            if (s1.charAt(i - 1) == s2.charAt(j - 1))
                arr[i][j] = arr[i - 1][j - 1] + 1;
            else
                arr[i][j] = Math.max(arr[i - 1][j], arr[i][j - 1]);
        }
    }
}

static Set<String> lcs(String s1, String s2, int len1, int len2) {
    if (len1 == 0 || len2 == 0) {
        Set<String> set = new HashSet<String>();
        set.add("");
        return set;
    }
    if (s1.charAt(len1 - 1) == s2.charAt(len2 - 1)) {
        Set<String> set = lcs(s1, s2, len1 - 1, len2 - 1);
        Set<String> set1 = new HashSet<>();
        for (String temp : set) {
            temp = temp + s1.charAt(len1 - 1);
            set1.add(temp);
        }
        return set1;
    } else {
        Set<String> set = new HashSet<>();
        Set<String> set1 = new HashSet<>();
        if (arr[len1 - 1][len2] >= arr[len1][len2 - 1]) {
            set = lcs(s1, s2, len1 - 1, len2);
        }
        if (arr[len1][len2 - 1] >= arr[len1 - 1][len2]) {
            set1 = lcs(s1, s2, len1, len2 - 1);
        }
        for (String temp : set) {
            set1.add(temp);
        }
        //System.out.println("In lcs" + set1);
        return set1;

    }
}


public static void main(String[] args) {
    String s1 = "bcab";
    String s2 = "abc";
    arr = new int[s1.length() + 1][s2.length() + 1];
    lcs(s1, s2);
    System.out.println(lcs(s1, s2, s1.length(), s2.length()));
}

Answer 2

在DP表中计算单元格时，请保留指向用于该结果的上一个单元格的后向指针。如果存在平局，请为所有绑定结果保留多个后退指针。然后使用后退指针沿着所有路径回溯路径。

Answer 3

这是一个注释的Java程序，用于查找所有可能的lcs。

import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class LongestCommonSubsequence {

    public static int[][] LCSmatrix(String X, String Y) {
        //we ignore the top most row and left most column in this matrix
        //so we add 1 and create a matrix with appropriate row and column size 
        int m = X.length() + 1, n = Y.length() + 1;

        int[][] c = new int[m][n];

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                //since we added 1 to row size and column size,
                // we substract 1 from i,j to find the char at that index
                if (X.charAt(i - 1) == Y.charAt(j - 1)) {
                    c[i][j] = c[i - 1][j - 1] + 1;
                } else if (c[i - 1][j] >= c[i][j - 1]) {
                    c[i][j] = c[i - 1][j];
                } else {
                    c[i][j] = c[i][j - 1];
                }
            }
        }
        printMatrix(c);
        return c;
    }

    public static void printMatrix(int[][] grid) {
        for (int r = 0; r < grid.length; r++) {
            for (int c = 0; c < grid[r].length; c++) {
                System.out.print(grid[r][c] + " ");
            }
            System.out.println();
        }
    }

    public static void allLCS(int[][] c, String X, String Y, int i, int j, Set<String> setLCS, String s) {
        //return when either of the string length is 0
        if (i == 0 || j == 0) {
            setLCS.add(s);
            return;
        }

        //if last characters are equal, they belong in lcs
        if (X.charAt(i - 1) == Y.charAt(j - 1)) {
            //prepend the char to lcs since, we are going backwards
            s = X.charAt(i - 1) + s;
            //continue finding lcs in substrings X.substring(0,i-1) and Y.substring(0,j-1)
            allLCS(c, X, Y, i - 1, j - 1, setLCS, s);
        } // if there is a tie in matrix cells, we backtrack in both ways,
        // else one way, which ever is greater
        else if (c[i - 1][j] == c[i][j - 1]) {
            //continue finding lcs in substring X.substring(0,i-1)
            allLCS(c, X, Y, i - 1, j, setLCS, s);
            //continue finding lcs in substring Y.substring(0,j-1)
            allLCS(c, X, Y, i, j - 1, setLCS, s);
        } else if (c[i - 1][j] > c[i][j - 1]) {
            allLCS(c, X, Y, i - 1, j, setLCS, s);
        } else {
            allLCS(c, X, Y, i, j - 1, setLCS, s);
        }

    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        System.out.println(" Enter String X and Y : ");

        String X = sc.next();
        String Y = sc.next();

        sc.close();

        Set<String> set = new HashSet<String>();
        allLCS(LCSmatrix(X, Y), X, Y, X.length(), Y.length(), set, "");

        System.out.println(set.toString());
    }

}

两个字符串的所有可能的LCS（最长公共子序列）

3 个答案: