題目連結：不同的子序列 - 力扣 (LeetCode)

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, “ACE” is a subsequence of “ABCDE” while “AEC” is not).

Example 1:

Input: S = "rabbbit", T = "rabbit"
Output: 3
Explanation:

As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)

rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^

Example 2:

Input: S = "babgbag", T = "bag"
Output: 5
Explanation:

As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)

babgbag
^^ ^
babgbag
^^    ^
babgbag
^    ^^
babgbag
  ^  ^^
babgbag
    ^^^
 

子序列是不用連續的（字串要連續）

這種題一般都需要轉換思考方式，用 S 中的字元，按順序匹配 T 中的字元。

$f\left[i\right]\left[j\right]$

• $S[i−1] \neq T[j−1]$，則 $S[i-1]$不匹配 $T[j-1]$， $f[i][j]=f[i−1][j]$
• $S[i−1]=T[j−1]$ ，則 $S[i−1]$既可以不匹配 $T[j−1]$，也可以匹配 $T[j−1]$，即 $f[i][j]=f[i−1][j]+f[i−1][j−1]$

邊界情況是T串不取時，匹配數都為1。 $f[][0] = 1$

class Solution {
public:
    int numDistinct(string s, string t) {
        int n = s.size(), m = t.size();
        vector<vector<int>>f(n + 1, vector<int>(m + 1, 0));
        for (int i = 0; i <= n; i ++ ) f[i][0] = 1;
        for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= m; j ++ )
            {
                f[i][j] = f[i-1][j];
                if (s[i-1] == t[j-1]) f[i][j] += f[i-1][j-1];
            }  
        return f[n][m];
    }
};