[LeetCode OJ] 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.

方法一：用回溯法实现，时间复杂度很高，空间复杂度低，对于小数据可以通过，对大数据会出现Time Limit Exceeded

int num=0;
void countnum(string S, string T) {
    if(T.size()==0)
    {
        num++;
        return;
     }
           
     for(int i=0; i<S.size(); i++)
     {
         if(S[i]==T[0])
         {
             string s2 = S.substr(i+1);
             string t2 = T.substr(1);
              countnum(s2, t2);
          }
              
     }
     return;
}

class Solution {
public:
    int numDistinct(string S, string T) {
        countnum(S, T);
        return num;
    }
};

方法二：用动态规划（DP）实现，需要的空间复杂度为O(N*M)，对于大数据也可以很快处理。

class Solution {
public:
    int numDistinct(string S, string T) {
        vector<vector<int> > num(S.size()+1,vector<int>(T.size()+1,0)); //num[i][j]表示T中的前j个字符构成的子字符串在S中的前i个字符中出现的次数,num[i][j]满足:
        S = " "+ S;                                                                                 //（1）若S[i]=T[j]，则num[i][j] = num[i-1][j]+num[i-1][j-1]；
        T = " "+ T;                                                                                 //（2）若S[i]!=T[j]，则num[i][j] = num[i-1][j]；
        num[0][0]=1;                                                                                //（3）若j>i，则num[i][j]=0。
        for(int i=1; i<S.size(); i++)
            for(int j=0; j<T.size(); j++)
            {
                if(j>i)
                {
                    num[i][j]=0;
                    break;
                }
                if(S[i]==T[j])
                    num[i][j] = num[i-1][j] + num[i-1][j-1];
                else
                    num[i][j] = num[i-1][j];
            }
            return num[S.size()-1][T.size()-1];

    }
};