Java LCS lcs(String X, String Y, int m, int n)

Description

lcs

License

Open Source License

Declaration

private static int lcs(String X, String Y, int m, int n) 

Method Source Code

//package com.java2s;
//License from project: Open Source License 

public class Main {
    private static int lcs(String X, String Y, int m, int n) {
        int[][] L = new int[m + 1][n + 1];

        // Following steps build L[m+1][n+1] in bottom up fashion. Note
        // that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]
        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                if (i == 0 || j == 0)
                    L[i][j] = 0;/* w ww  . java  2s  .co  m*/
                else if (X.charAt(i - 1) == Y.charAt(j - 1))
                    L[i][j] = L[i - 1][j - 1] + 1;
                else
                    L[i][j] = Math.max(L[i - 1][j], L[i][j - 1]);
            }
        }

        // Print the table
        System.out.print("  ");
        for (int j = 1; j <= n; j++) {
            System.out.print(" " + Y.charAt(j - 1) + " ");
        }
        System.out.println();

        for (int i = 1; i <= m; i++) {
            System.out.print(X.charAt(i - 1) + " ");
            for (int j = 1; j <= n; j++) {
                String formatted = String.format("%02d", L[i][j]);
                System.out.print(formatted + " ");
            }
            System.out.println();
        }

        // Following code is used to print LCS
        int index = L[m][n];
        int temp = index;

        // Create a character array to store the lcs string
        char[] lcs = new char[index + 1];
        lcs[index] = ' '; // Set the terminating character

        // Start from the right-most-bottom-most corner and
        // one by one store characters in lcs[]
        int i = m;
        int j = n;
        while (i > 0 && j > 0) {
            // If current character in X[] and Y are same, then
            // current character is part of LCS
            if (X.charAt(i - 1) == Y.charAt(j - 1)) {
                // Put current character in result
                lcs[index - 1] = X.charAt(i - 1);

                // reduce values of i, j and index
                i--;
                j--;
                index--;
            }

            // If not same, then find the larger of two and
            // go in the direction of larger value
            else if (L[i - 1][j] > L[i][j - 1])
                i--;
            else
                j--;
        }

        // Print the lcs
        System.out.print("LCS of \n" + X + "\n  and \n" + Y + "\n  is \n");
        for (int k = 0; k <= temp; k++)
            System.out.print(lcs[k]);

        return L[m][n];
    }
}

Related

1. lcs(String arg0, String arg1)
2. LCS(String input1, String input2)
3. lcs(String s1, int s1min, int s1max, String s2, int s2min, int s2max)
4. lcs(String s1, String s2)
5. lcs(String str1, String str2)
6. lcs1(int[] A, int[] B)
7. lcs2(int[] A, int[] B, int m, int n)
8. lcs3(int[] A, int[] B)
9. lcs4(int[] A, int[] B)