经典DP之两字符串最长公共子序列

转载请注明作者与出处：franciscolv http://www.cnblogs.com/franciscolv/archive/2011/11/20/2255858.html

分为连续子串和非连续子串。

package SortSet;

/**
 * 最长公共子序列 连续 非连续
 * 
 * @author franciscolv
 * 
 */
public class LCS {
    private static char[] result;
    private static int posj;

    public static void main(String[] args) {
        String s = "ABCDEFGABCAGD";
        String t = "ABCDFGABCDGD";
        System.out.println("非连续最长公共子串为长度为："
                + lcs(s.toCharArray(), t.toCharArray()));
        System.out.println("非连续最长公共子串为：");
        for (char c : result)
            System.out.print(c);
        System.out.println();
        int k = lcs2(s.toCharArray(), t.toCharArray());
        System.out.println("连续最长公共子串为长度为：" + k);
        System.out.println("连续最长公共子串为：");
        for (int i = posj - k + 1; i < posj + 1; i++)
            System.out.print(t.charAt(i));
        System.out.println();
    }

    /**
     * o(n*n) 最长连续公共子序列
     * 
     * @param c
     * @param b
     * @return
     */
    public static int lcs2(char[] c, char[] b) {
        int[][] m = new int[c.length + 1][b.length + 1];
        for (int i = 1; i <= c.length; i++) {
            for (int j = 1; j <= b.length; j++) {
                if (c[i - 1] == b[j - 1]) {
                    m[i][j] = m[i - 1][j - 1] + 1;
                } else
                    m[i][j] = 0;

            }
        }
        return max(m, c.length + 1, b.length + 1);
    }

    private static int max(int[][] m, int len1, int len2) {
        int max = -1;
        for (int i = 1; i < len1; i++)
            for (int j = 1; j < len2; j++)
                if (m[i][j] > max) {
                    max = m[i][j];
                    posj = j - 1;
                }

        return max;
    }

    /**
     * 非连续最长公共子序列 o(n*n)
     * 
     * @param c
     * @param b
     * @return
     */
    public static int lcs(char[] c, char[] b) {
        int[][] m = new int[c.length + 1][b.length + 1];
        int[][] s = new int[c.length + 1][b.length + 1];
        for (int i = 1; i <= c.length; i++) {
            for (int j = 1; j <= b.length; j++) {
                if (c[i - 1] == b[j - 1]) {
                    m[i][j] = m[i - 1][j - 1] + 1;
                } else {
                    if (m[i - 1][j] > m[i][j - 1]) {
                        m[i][j] = m[i - 1][j];
                        s[i][j] = -1;
                    } else {
                        m[i][j] = m[i][j - 1];
                        s[i][j] = 1;
                    }
                }
            }
        }
        int p = c.length, q = b.length, k = m[p][q];
        if (k > 0) {
            result = new char[k];
            while (p > 0 && q > 0) {
                switch (s[p][q]) {
                case -1:
                    p--;
                    break;
                case 0:
                    result[--k] = c[p - 1];
                    p--;
                    q--;
                    break;

                case 1:
                    q--;
                    break;
                }
            }
        }
        return m[c.length][b.length];
    }

}