[LeetCode] Distinct Subsequences

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.

1. 递归比较，超时！！！

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;        match(S,T,-1,-1,result);        return result;    }        void match(string S, string T, int lastS, int lastT, int &result)    {        if(lastT==T.length()-1)         {            result++;            return;        }                for(int i=lastS+1;i<S.length();i++)        {            if(T[lastT+1]==S[i])            {                match(S,T,i,lastT+1,result);            }        }            }};

2. dynamic programming: 

t[i][j]=the number of distinct subsequences of string T of length i in string  S of length j. So finally the program returns t[t.length][s.length]

If T[i]!=S[j], then t[i][j]=the number of distinct subsequences of string T of length i in string  S of length j-1=t[i][j-1] (the left item in the matrix)

If T[i]==S[j], then t[i][j]=the number of distinct subsequences of string T of length i in string  S of length j-1 + the number of distinct subsequences of string T of length i-1 in string  S of length j-1 = t[i][j-1] + t[i-1][j-1] (the left item in the matrix+the left up item)

class Solution {public:    int numDistinct(string S, string T) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int>> t(T.length()+1,vector<int>(S.length()+1));                for(int i=0;i<=T.length();i++) t[i][0]=0;        for(int i=0;i<=S.length();i++) t[0][i]=0;                for(int i=1;i<=S.length();i++)        {            if(T[0]==S[i-1])                t[1][i]=t[1][i-1]+1;            else                t[1][i]=t[1][i-1];            }                        for(int i=2;i<=T.length();i++)        {            for(int j=1; j<=S.length();j++)            {                if(T[i-1]==S[j-1])                    t[i][j]=t[i-1][j-1]+t[i][j-1];                else                    t[i][j]=t[i][j-1];            }        }                return t[T.length()][S.length()];    }};