Distinct Subsequences

-----QUESTION-----

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

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).

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

-----SOLUTION-----

class Solution {
public:
    int numDistinct(string S, string T) {
        int tLen = T.length();
        int sLen = S.length();
        if(sLen == 0) return 0;
        vector<vector<int>> dp(sLen,vector<int>(tLen));  
       
        int i=0;
        int j=0;
        if(S[0]==T[0]) dp[0][0] = 1;
        else dp[0][0]=0;
        for(i = 1; i < tLen; i++)
            dp[0][i]=0;
        for(i = 1 ; i < sLen ;i++)  
            if(S[i]==T[0])  
                dp[i][0] = dp[i-1][0] + 1;  
            else  
                dp[i][0] = dp[i-1][0]; 
                
        for(int i = 1;  i < sLen ; i++)  
            for(int j = 1; j < tLen ; j++)  
                if(S[i]==T[j])  
                    dp[i][j] = dp[i-1][j-1] + dp[i-1][j];  
                else  
                    dp[i][j] = dp[i-1][j];  
       return dp[sLen-1][tLen-1];  
    }
};

下面做法在Judge Large时，time limit exceeded

class Solution {
public:
    int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int result = 0;
        solve(S,T,0,0,result);
        return result;     
    }
    void solve(const string & S, const string & T, int sPos, int tPos, int & result)
    {
        int tLen = T.length();
        int sLen = S.length();
       
        for (int i = tPos; i< tLen; i++)
        {
            if(sPos >= sLen)
                return;
            while(S[sPos]!=T[i])
            {
                sPos++;
                if(sPos >= sLen)
                    return;
            }          
            sPos++;    
            solve(S,T,sPos,i,result);
        }
        result ++;
    }
};