115. Distinct Subsequences *HARD* -- 字符串不连续匹配

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.

class Solution {
public:
    int numDistinct(string s, string t) {
        int ls = s.length(), lt = t.length(), i, j;
        vector<vector<int>> dp(lt+1, vector<int>(ls+1, 0));
        for(i = 0; i <= ls; i++)
            dp[0][i] = 1;
        for(i = 1; i <= lt; i++)
        {
            for(j = i; j <= ls; 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];
                }
            }
        }
        return dp[lt][ls];
    }
};

注意：

“bbb”和“”的匹配结果为1。所以第一行都为1。