最長公共子序列LCS（動態規劃）

public class LCS
{
    public static void main(String[] args)
    {
     
        String[] x = {"", "U","B", "S", "B", "T", "H", "C"};
        String[] y = {"", "B", "U", "S", "A", "E", "T", "A", "C"};
        int[][] b = getLength(x, y);
        Display(b, x, x.length-1, y.length-1);
    }
    /*
     * @param x
     * @param y
     * @return 返回一個記錄決定搜索的方向的數組
     */
    public static int[][] getLength(String[] x, String[] y)
    {
    //這裏沒有把第一行第一列置爲0
        int[][] b = new int[x.length][y.length];
        int[][] c = new int[x.length][y.length];
        for(int i=1; i<x.length; i++)
        {
            for(int j=1; j<y.length; j++)
            {
                //對應第一個性質
                if( x[i] == y[j])
                {
                    c[i][j] = c[i-1][j-1] + 1;
                    b[i][j] = 1;
                }
                //對應第二或者第三個性質
                else if(c[i-1][j] >= c[i][j-1])
                {
                    c[i][j] = c[i-1][j];
                    b[i][j] = 0;
                }
                //對應第二或者第三個性質
                else
                {
                    c[i][j] = c[i][j-1];
                    b[i][j] = -1;
                }
            }
        }
        return b;
    }
    //回溯的基本實現，採取遞歸的方式
    public static void Display(int[][] b, String[] x, int i, int j)
    {
        if(i == 0 || j == 0)
            return;
        if(b[i][j] == 1)
        {
            Display(b, x, i-1, j-1);
            System.out.print(x[i]);
        }
        else if(b[i][j] == 0)
        {
            Display(b, x, i-1, j);
        }
        else if(b[i][j] == -1)
        {
            Display(b, x, i, j-1);
        }
    }
}