首先下意识想到的就是直接遍历两个字符串，相同的就存储起来，最终得到结果呗。

private static String LCS1(String str1, String str2) {
		// TODO Auto-generated method stub
		StringBuilder sb = new StringBuilder();
		int len1,len2,i,j;
		sb.delete(0, sb.length());
		len1 = str1.length();
		len2 = str2.length();
		int tmp=0;
		for(j=0;j<len1;j++){
			for(i=tmp;i<len2;i++){
				if(str1.charAt(j)==str2.charAt(i)){
					sb.append(str2.charAt(i));
					tmp = i+1;
					break;
				}
			}
		}
		return sb.toString();
	}

备份的递归方法，觉得最后的重构还是很好玩的，而且这种算法通用性比较强嘛，就实现一下呗。

package 动态规划;

import java.util.Scanner;

public class 最长公共子序列_备忘机制 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner cin = new Scanner(System.in);
		String str1,str2;
		while(cin.hasNext()){
			str1 = cin.nextLine();
			str2 = cin.nextLine();
			int n1 = str1.length();
			int n2 = str2.length();
			int[][] c = new int[n1+1][n2+1];
			for(int i=0;i<=n1;i++)
				for(int j=0;j<=n2;j++){
					if(i==0||j==0)
						c[i][j] = 0;
					else
						c[i][j] = Integer.MIN_VALUE;
				}
			int[][] b = new int[n1+1][n2+1];
			System.out.println(up_to_bottom_LCS(str1,str2,c,b,n1,n2));
			printLCS(b, str1, n1,n2);
			System.out.println();
		}
	}

	private static void printLCS(int[][] b, String str1, int n1, int n2) {
		// TODO Auto-generated method stub
		if(n1==0||n2==0)
			return;
		if(b[n1][n2]==2){
			printLCS(b, str1, n1-1, n2-1);
			System.out.print(str1.charAt(n1-1));
		}else if(b[n1][n2]==-1)
			printLCS(b, str1, n1-1, n2);
		else
			printLCS(b, str1, n1, n2-1);
	}

	private static int up_to_bottom_LCS(String str1, String str2, int[][] c,
			int[][] b, int n1, int n2) {
		// TODO Auto-generated method stub
		if(n1==0||n2==0)
			return 0;
		if(c[n1][n2]>Integer.MIN_VALUE)
			return c[n1][n2];
		if(str1.charAt(n1-1)==str2.charAt(n2-1)){
			c[n1][n2] = up_to_bottom_LCS(str1, str2, c, b, n1-1, n2-1)+1;
			b[n1][n2] = 2;
		}else{
			if(up_to_bottom_LCS(str1, str2, c, b, n1-1, n2)>up_to_bottom_LCS(str1, str2, c, b, n1, n2-1)){
				c[n1][n2] = up_to_bottom_LCS(str1, str2, c, b, n1-1, n2);
				b[n1][n2] = -1;
			}else{
				c[n1][n2] = up_to_bottom_LCS(str1, str2, c, b, n1, n2-1);
				b[n1][n2] = 1;
			}
		}
		return c[n1][n2];
	}
}