[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.

动态规划！

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

其实最开始的想法是dfs，果段超时。比起动规来还是递归好写一点，动规还不太熟练，得加油啊。

class Solution {
public:
    void dfs(string &S, int idxs, string &T, int idxt, int &res) {
        if (idxt == T.length()) {
            ++res;
            return;
        }
        if (idxs == S.length()) {
            return;
        }
        if (S[idxs] == T[idxt]) {
            dfs(S, idxs + 1, T, idxt + 1, res);
        } 
        dfs(S, idxs + 1, T, idxt, res);
    }
    
    int numDistinct(string S, string T) {
        int res = 0;
        dfs(S, 0, T, 0, res);
        return res;
    }
};