DP3_最长公共子序列

 1 using System;
 2 namespace ConsoleApplication2
 3 {
 4     public class Program
 5     {
 6         static int[,] martix;
 7 
 8         static string str1 = "cnblogs";
 9         static string str2 = "belong";
10 
11         static void Main(string[] args)
12         {
13             martix = new int[str1.Length + 1, str2.Length + 1];
14 
15             LCS(str1, str2);
16 
17             //只要拿出矩阵最后一个位置的数字即可
18             Console.WriteLine("当前最大公共子序列的长度为:{0}", martix[str1.Length, str2.Length]);
19 
20             Console.Read();
21         }
22 
23         static void LCS(string str1, string str2)
24         {
25             //初始化边界，过滤掉0的情况
26             for (int i = 0; i <= str1.Length; i++)
27                 martix[i, 0] = 0;
28 
29             for (int j = 0; j <= str2.Length; j++)
30                 martix[0, j] = 0;
31 
32             //填充矩阵
33             for (int i = 1; i <= str1.Length; i++)
34             {
35                 for (int j = 1; j <= str2.Length; j++)
36                 {
37                     //相等的情况
38                     if (str1[i - 1] == str2[j - 1])
39                     {
40                         martix[i, j] = martix[i - 1, j - 1] + 1;
41                     }
42                     else
43                     {
44                         //比较“左边”和“上边“，根据其max来填充
45                         if (martix[i - 1, j] >= martix[i, j - 1])
46                             martix[i, j] = martix[i - 1, j];
47                         else
48                             martix[i, j] = martix[i, j - 1];
49                     }
50                 }
51             }
52         }
53     }
54 }

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int MAX(int a, int b)
 6 {
 7     return a>b?a:b;
 8 }
 9 int f(char* x, char* y)
10 {
11     if(strlen(x)==0) return 0;
12     if(strlen(y)==0) return 0;
13     if(*x == *y) return f(x+1,y+1)+1;
14     return  MAX(f(x,y+1),f(x+1,y));
15 }
16 int main()
17 {
18     printf("%d\n", f("ac","abcd")); //2
19     printf("%d\n", f("acebbcde1133","xya33bc11de")); //5
20     return 0;
21 }