Longest common subsequence __java achieve

Key Code:

for(int i=1;i<=x.length;i++){
    for(int j=1;j<=y.length;j++){
        if(x[i-1]==y[j-1]){
            c[i][j] = c[i-1][j-1]+1;
            b[i][j] = 1;
        }else{
            if(c[i-1][j]>=c[i][j-1]){
                c[i][j] = c[i-1][j];
                b[i][j] = 2;
            }else{
                c[i][j] = c[i][j-1];
                b[i][j] = 3;
            }
        }
    }
}

Above, i = 0 or j = 0, i.e. X or Y is an empty sequence, c [i] [j] = 0.

 

Complete code:

{class ZCGGZXL public 
    / ** 
     * longest common subsequence 
     * 
     * Question: 
     * Given two sequences X = {x1, x2, ... , xn} and Y = {y1, y2, ... ,, yn }, find the longest common subsequence of X and Y. 
     * 
     * Test input: 
     * ABCBDAB 
     * BDCABA 
     * Test Output: 
     *. 4 
     * BCBA 
     * / 

    public static void main (String args []) { 
        char [] X = { 'A', 'B', 'C', 'B ',' D ',' A ',' B '}; 
        char [] Y = {' B ',' D ',' C ',' A ',' B ',' A '}; 
        int [] [ ] B = new new int [x.length +. 1] [y.length +. 1]; 
        int [] [] = lcsLength C (X, Y, B); 
        System.out.println (C [x.length] [Y .

    / ** 
     * Input: x sequence, y sequence 
     * Output: b array, return the array c. 
     * C [i] [j] the length of the longest common subsequence of Xi and Yj stored in the 
     value * b [i] [j] recorded c [i] [j] which is a sub-problem solution obtained in to use when constructing the longest common subsequence. 
     @Return * 
     * / 
    public static int [] [] lcsLength (char [] X, char [] Y, int [] [] B) { 
        // initialize an array C 
        int [] [] = C new new int [x.length +1] [y.length + 1]; // 0 kept empty sequence 
        for (int I = 0; I <c.length; I ++) { 
            for (int J = 0; J <C [0] .length; J ++ ) { 
                C [I] [J] = 0; 
            } 
        } 
        // start planning 
        for (int I =. 1; I <= x.length; I ++) { 
            for (int. 1 = J; J <= y.length; J ++ ) { 
                IF (X [I-. 1] == Y [J-. 1]) {
                    C [I] [J] = C [. 1-I] [J-. 1] + 1'd; 
        if (b [i] [j ] == 1) {
                    B [I] [J] =. 1; 
                } the else { 
                    IF (C [I-. 1] [J]> = [I] C [J-. 1]) { 
                        C [I] [J] = C [I-. 1 ] [J]; 
                        B [I] [J] = 2; 
                    } the else { 
                        C [I] [J] = C [I] [J-. 1]; 
                        B [I] [J] =. 3; 
                    } 
                } 
            } 
        } 
        C return; 
    } 

    public static void LCS (I int, int J, char [] X, int [] [] B) { 
        // end condition 
        IF (I == 0 || J == 0) return; 

        // Analyzing b [i] [j] into a different branch 
            LCS (-I. 1,. 1-J, X, B); 
            of System.out.print (X [I-. 1]);
        }else{
            if(b[i][j]==2) lcs(i-1,j,x,b);
            else lcs(i,j-1,x,b);
        }
    }
}