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.

思路：

DP。

用r[i][j]表示S的前i的子串包含多少T的前j的子串。

如果S[i] == T[j], r[i][j] = r[i-1][j] + r[i-1][j-1]

如果S[i] !=  T[j], r[i][j] = r[i-1][j]。

代码：

int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int col = S.length() + 1;
        int row = T.length() + 1;
        int** dp = new int*[row];
        for(int i = 0; i < row; ++i)
            dp[i] = new int[col];
        
        for(int i = 0; i < row; ++i)
            dp[i][0] = 0;
        for(int j = 0; j < col; ++j)
            dp[0][j] = 1;
        
        for(int i = 1; i < row; ++i)
            for(int j = 1; j < col; ++j)
                if(T[i-1] == S[j-1]) dp[i][j] = dp[i-1][j-1] + dp[i][j-1];
                else dp[i][j] = dp[i][j-1];
                
        int tmp = dp[row-1][col-1];
        
        for(int i = 0; i < row; ++i)
            delete[] dp[i];
        delete[] dp;
        return tmp;
    }

 第二次

int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int ls = S.length(), lt = T.length();
        int res[ls+1][lt+1];
        int i,j;
        for(i = 0; i <= lt; i++)
            res[0][i] = 0;
        for(i = 0; i <= ls; i++)
            res[i][0] = 1;
        for(i = 1; i <= ls; i++){
            for(j = 1; j <= lt; j++){
                if(S[i-1] == T[j-1])
                    res[i][j] = res[i-1][j-1]+res[i-1][j];
                else
                    res[i][j] = res[i-1][j];
            }
        }
        return res[ls][lt];
    }