Dynamic Programming - Part2

实现如下：

public static void main(String[] args) {

        String squence1 = "ABCBDAB";
        String squence2 = "BDCABC";

        int len1 = squence1.length(), len2 = squence2.length();
        if (len1 <= 0 || len2 <= 0) {
            System.out.println("empty squence!");
            return;
        }
        int[][] allCommonSubLen = new int[len1+1][len2+1];
        //print all sub-squences
        String subSquence = "";
        //初始化i=0, j=0
        for (int i = 0; i < len1; i++) {
            allCommonSubLen[i][0] = 0;
        }
        for (int j = 0; j < len2; j++) {
            allCommonSubLen[0][j] = 0;
        }
        //迭代关系
        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                char c1 = squence1.charAt(i-1);
                char c2 = squence2.charAt(j-1);
                if (c1 == c2) {
                    //subSquence += c1;
                    allCommonSubLen[i][j] = allCommonSubLen[i-1][j-1] + 1;
                    //System.out.println("i="+i+" ; j="+j+" ; subSquence="+subSquence);
                } else {
                    allCommonSubLen[i][j] = Math.max(allCommonSubLen[i-1][j], allCommonSubLen[i][j-1]);
                    if (allCommonSubLen[i][j] == allCommonSubLen[i-1][j]) {
                        //subSquence += c2;
                        //System.out.println("i="+i+" ; j="+j+" ; subSquence="+subSquence);
                    } else {
                        //subSquence += c1;
                        //System.out.println("i="+i+" ; j="+j+" ; subSquence="+subSquence);
                    }
                }
            }
        }
        //System.out.println(allCommonSubLen[len1][len2]);
    }