动态规划--最长公共子串

最长公共子串也是一个动态规划的问题，先求出子问题的解，然后才能求出最优解。

首先我们来看设X = <x₁, x₂, ..., x_m>, Y= <y₁, y₂, ..., y_n>，设C[i][j]为串 X_i 和 Y_j 的最长公共子串长度，则

• C[i][j]  =  C[i-1][j-1] +1,  X_i == Y_j
• C[i][j]  =  0,  X_i != Y_j

申请一个m*n的数组，同时计算出该数组对应位置的数值，找出最大值，则为X 和 Y最长公共子串。

代码如下：

 1 // LCS(SubString).cpp : 定义控制台应用程序的入口点。
 2 //
 3
 4 #include <iostream>
 5 #include <string>
 6 using namespace std;
 7
 8
 9 int LengthOfLongestSubString(const string &X, const string &Y)
10 {
11     // 使元素从下标0开始存储，方便数组访问
12     string strX = " " + X, strY = " " +Y;
13     int m = strX.size(), n = strY.size();
14     int maxLength = 0;
15
16     int **c = new int*[m];
17
18     for (int i = 0; i < m; ++i)
19         c[i] = new int[n];
20     // init the cases that i = 0 and j = 0
21     for (int i = 0; i < m; ++i)
22         c[i][0] = 0;
23     for (int i = 0; i < n; ++i)
24         c[0][i] = 0;
25
26     // Calculate the array C
27     for (int i = 1; i < m; i++)
28         for (int j = 1; j < n; ++j)
29         {
30             if (X[i] == Y[j])
31             {
32                 c[i][j] = c[i - 1][j - 1] + 1;
33                 if (c[i][j] > maxLength)
34                     maxLength = c[i][j];
35             }
36             else
37                 c[i][j] = 0;
38         }
39     // Free the memory
40     for (int i = 0; i < m; ++i)
41         delete c[i];
42     delete []c;
43
44     return maxLength;
45 }
46
47
48 int main(int argc, char** argv)
49 {
50     const string X = "ABCDEFDBEFG";
51     const string Y = "DKGDBCDEFAB";
52
53     int Length = LengthOfLongestSubString(X, Y);
54     cout << "The max Length is :" << Length << endl;
55
56     return 0;
57 }