題目鏈接
tag:

• Hard；

question：
??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
^^^

思路：
??看到有關字符串的子序列或者配準類的問題，首先應該考慮的就是用動態(tài)規(guī)劃Dynamic Programming來求解，這個應成為條件反射。而所有DP問題的核心就是找出遞推公式，想這道題就是遞推一個二維的dp數(shù)組，下面我們從題目中給的例子來分析，這個二維dp數(shù)組應為：

? r a b b b i t
? 1 1 1 1 1 1 11
r 0 1 1 1 1 1 1 1
a 0 0 1 1 1 1 1 1
b 0 0 0 1 2 3 3 3
b 0 0 0 0 1 3 3 3
i 0 0 0 0 0 0 3 3
t 0 0 0 0 0 0 0 3

?? 首先，若原字符串和子序列都為空時，返回1，因為空串也是空串的一個子序列。若原字符串不為空，而子序列為空，也返回1，因為空串也是任意字符串的一個子序列。而當原字符串為空，子序列不為空時，返回0，因為非空字符串不能當空字符串的子序列。理清這些，二維數(shù)組dp的邊緣便可以初始化了，下面只要找出遞推式，就可以更新整個dp數(shù)組了。我們通過觀察上面的二維數(shù)組可以發(fā)現(xiàn)，當更新到dp[i][j]時，dp[i][j] >= dp[i][j - 1] 總是成立，再進一步觀察發(fā)現(xiàn)，當 T[i - 1] == S[j - 1] 時，dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1]，若不等， dp[i][j] = dp[i][j - 1]，所以，綜合以上，遞推式為：

??dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)

代碼如下：

class Solution {
public:
    int numDistinct(string s, string t) {
        if (s.size() < t.size())
            return 0;
        int n1=t.size(), n2=s.size();
        long dp[n1+1][n2+1];
        
        for (int i=0; i<=n1; ++i)
            dp[i][0] = 0;
        for (int i=0; i<=n2; ++i)
            dp[0][i] = 1;
        
        for (int i=1; i<=n1; ++i) {
            for (int j=1; j<=n2; ++j) {
                if (t[i-1] == s[j-1])
                    dp[i][j] = dp[i][j-1] + dp[i-1][j-1];
                else
                    dp[i][j] = dp[i][j-1];
            }
        }
        return dp[n1][n2];
    }
};