08_最长连续子串

结合这两篇文章，使用dp求解，便于理解

http://www.cnblogs.com/en-heng/p/3963803.html

http://www.cnblogs.com/xudong-bupt/archive/2013/03/15/2959039.html

状态转移方程

矩阵求解

import java.util.Arrays;

public class Main {

    public static void main(String[] args) {
        lcs("ABCBDAB", "BDCABA");
    }

    public static int lcs(String str1, String str2) {
        int len1 = str1.length();
        int len2 = str2.length();
        int result = 0; // 记录最长公共子串长度
        int c[][] = new int[len1 + 1][len2 + 1];
        for (int i = 0; i <= len1; i++) {
            for (int j = 0; j <= len2; j++) {
                ////第一行和第一列全部初始化为0.类似于迷宫问题外墙的功能
                if (i == 0 || j == 0) {
                    c[i][j] = 0;
                } 
                 //拿出str1的某个字符，遍历str2的字符，相等的话就斜对角加1
                else if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
                    c[i][j] = c[i - 1][j - 1] + 1;
                    result = Math.max(c[i][j], result);
                }
                //不相等的话就重新置为0
                else {
                    c[i][j] = 0;
                }
            }
        }
        for (int i = 0; i < len1 + 1; i++) {
            System.out.println(Arrays.toString(c[i]));
        }
        return result;
    }

}

矩阵求解过程示意图