## 二、LCS算法

### 2.1 生成矩阵

i和j分别从1开始，i++，j++循环

• 如果str1[i] == str2[j]，则L[i,j] = L[i - 1, j -1] + 1

• 如果str1[i] != str2[j]，则L[i,j] = max{L[i,j - 1]，L[i - 1, j]}

### 2.2 计算公共子序列

ij分别从str1_lenstr2_len开始，递减循环直到i = 0，j = 0

• 如果str1[i-1] == str2[j-1]，则将str[i]字符插入到子序列中，i--，j--

• 如果str1[i-1] != str[j-1]，则比较L[i,j-1]L[i-1,j]L[i,j-1]大，则j--，否则i--；（如果相等，则任选一个

### 2.3 逆序存放公共子序列

step2得到的公共子序列是从后往前获得的，需要逆序存放或输出

## 四、牛刀小试

POJ 1458 Common Subsequence

Description
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, …, xm > another sequence Z = < z1, z2, …, zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, …, ik > of indices of X such that for all j = 1,2,…,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.

Input
The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output
For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

Sample Input

abcfbc abfcab
programming contest
abcd mnp

Sample Output

4
2
0

Code