计算两个字符串的公共子序列

给定两个字符串S和Q 。任务是统计S和T中公共子序列的个数。
例子：

Input : S = “ajblqcpdz”, T = “aefcnbtdi” 
Output : 11 
Common subsequences are : { “a”, “b”, “c”, “d”, “ab”, “bd”, “ad”, “ac”, “cd”, “abd”, “acd” }
Input : S = “a”, T = “ab” 
Output : 1

为了找到两个字符串的公共子序列的数量，比如说 S 和 T，我们通过定义一个二维数组dp[][] 来使用动态规划，其中 dp[i][j] 是字符串S[ 中的公共子序列的数量0…i-1] 和 T[0….j-1]。
现在，我们可以将 dp[i][j] 定义为
= dp[i][j-1] + dp[i-1][j] + 1，当 S[i-1] 等于 T[j-1]
这是因为当 S[i-1] == S[j-1] 时，使用上述事实，所有先前的公共子序列都加倍了，因为它们再追加一个字符。 dp[i][j-1] 和 dp[i-1][j] 都包含 dp[i-1][j-1] ，因此在我们的循环中它被添加了两次，它负责将计数加倍所有以前的公共子序列。在重复中加 1 是针对最新的字符匹配：由 s1[i-1] 和 s2[j-1] 组成的公共子序列
= dp[i-1][j] + dp[i][j-1] – dp[i-1][j-1]，当 S[i-1] 不等于 T[j-1]
在这里，我们减去 dp[i-1][j-1] 一次，因为它同时存在于 dp[i][j – 1] 和 dp[i – 1][j] 中并且被添加了两次。
下面是这个方法的实现：

C++

// C++ program to count common subsequence in two strings
#include 
using namespace std;
 
// return the number of common subsequence in
// two strings
int CommomSubsequencesCount(string s, string t)
{
    int n1 = s.length();
    int n2 = t.length();
    int dp[n1+1][n2+1];
 
    for (int i = 0; i <= n1; i++) {
        for (int j = 0; j <= n2; j++) {
            dp[i][j] = 0;
        }
    }
 
    // for each character of S
    for (int i = 1; i <= n1; i++) {
 
        // for each character in T
        for (int j = 1; j <= n2; j++) {
 
            // if character are same in both
            // the string
            if (s[i - 1] == t[j - 1])
                dp[i][j] = 1 + dp[i][j - 1] + dp[i - 1][j];           
            else
                dp[i][j] = dp[i][j - 1] + dp[i - 1][j] -
                                        dp[i - 1][j - 1];           
        }
    }
 
    return dp[n1][n2];
}
 
// Driver Program
int main()
{
    string s = "ajblqcpdz";
    string t = "aefcnbtdi";
 
    cout << CommomSubsequencesCount(s, t) << endl;
    return 0;
}

Java

// Java program to count common subsequence in two strings
public class GFG {
     
    // return the number of common subsequence in
    // two strings
    static int CommomSubsequencesCount(String s, String t)
    {
        int n1 = s.length();
        int n2 = t.length();
        int dp[][] = new int [n1+1][n2+1];
        char ch1,ch2 ;
       
        for (int i = 0; i <= n1; i++) {
            for (int j = 0; j <= n2; j++) {
                dp[i][j] = 0;
            }
        }
       
        // for each character of S
        for (int i = 1; i <= n1; i++) {
       
            // for each character in T
            for (int j = 1; j <= n2; j++) {
                 
                ch1 = s.charAt(i - 1);
                ch2 = t.charAt(j - 1);
                 
                // if character are same in both 
                // the string                
                if (ch1 == ch2) 
                    dp[i][j] = 1 + dp[i][j - 1] + dp[i - 1][j];            
                else
                    dp[i][j] = dp[i][j - 1] + dp[i - 1][j] - 
                                            dp[i - 1][j - 1];            
            }
        }
       
        return dp[n1][n2];
    }
    // Driver code
    public static void main (String args[]){
         
          String s = "ajblqcpdz";
          String t = "aefcnbtdi";
           
        System.out.println(CommomSubsequencesCount(s, t));
           
    }
 
// This code is contributed by ANKITRAI1
}

Python3

# Python3 program to count common
# subsequence in two strings
 
# return the number of common subsequence
# in two strings
def CommomSubsequencesCount(s, t):
 
    n1 = len(s)
    n2 = len(t)
    dp = [[0 for i in range(n2 + 1)]
             for i in range(n1 + 1)]
 
    # for each character of S
    for i in range(1, n1 + 1):
 
        # for each character in T
        for j in range(1, n2 + 1):
 
            # if character are same in both
            # the string
            if (s[i - 1] == t[j - 1]):
                dp[i][j] = (1 + dp[i][j - 1] +
                                dp[i - 1][j])        
            else:
                dp[i][j] = (dp[i][j - 1] + dp[i - 1][j] -
                            dp[i - 1][j - 1])        
         
    return dp[n1][n2]
 
# Driver Code
s = "ajblqcpdz"
t = "aefcnbtdi"
 
print(CommomSubsequencesCount(s, t))
 
# This code is contributed by Mohit Kumar

C#

// C# program to count common
// subsequence in two strings
using System;
 
class GFG
{
 
// return the number of common
// subsequence in two strings
static int CommomSubsequencesCount(string s,
                                   string t)
{
    int n1 = s.Length;
    int n2 = t.Length;
    int[,] dp = new int [n1 + 1, n2 + 1];
     
    for (int i = 0; i <= n1; i++)
    {
        for (int j = 0; j <= n2; j++)
        {
            dp[i, j] = 0;
        }
    }
     
    // for each character of S
    for (int i = 1; i <= n1; i++)
    {
     
        // for each character in T
        for (int j = 1; j <= n2; j++)
        {
             
            // if character are same in
            // both the string                
            if (s[i - 1] == t[j - 1])
                dp[i, j] = 1 + dp[i, j - 1] +
                               dp[i - 1, j];            
            else
                dp[i, j] = dp[i, j - 1] +
                           dp[i - 1, j] -
                           dp[i - 1, j - 1];            
        }
    }
     
    return dp[n1, n2];
}
 
// Driver code
public static void Main ()
{
    string s = "ajblqcpdz";
    string t = "aefcnbtdi";
         
    Console.Write(CommomSubsequencesCount(s, t));
}
}
 
// This code is contributed
// by ChitraNayal

Javascript

 
// Javscript program to count common subsequence in two strings
 
// return the number of common subsequence in
// two strings
function CommomSubsequencesCount(s, t)
{
    var n1 = s.length;
    var n2 = t.length;
    var dp = Array.from(Array(n1+1), ()=> Array(n2+1));
 
    for (var i = 0; i <= n1; i++) {
        for (var j = 0; j <= n2; j++) {
            dp[i][j] = 0;
        }
    }
 
    // for each character of S
    for (var i = 1; i <= n1; i++) {
 
        // for each character in T
        for (var j = 1; j <= n2; j++) {
 
            // if character are same in both
            // the string
            if (s[i - 1] == t[j - 1])
                dp[i][j] = 1 + dp[i][j - 1] + dp[i - 1][j];           
            else
                dp[i][j] = dp[i][j - 1] + dp[i - 1][j] -
                                        dp[i - 1][j - 1];           
        }
    }
 
    return dp[n1][n2];
}
 
// Driver Program
var s = "ajblqcpdz";
var t = "aefcnbtdi";
document.write( CommomSubsequencesCount(s, t));
 
 

输出：

11

时间复杂度： O(n1 * n2)
辅助空间： O(n1 * n2)
来源： StackOverflow

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