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[i][j]表示 # of T[0...j-1] in S[0...i-1] (dp[0][0]表示s=NULL，t=NULL的情况）

如果S[i]！=T[j]，那么dp[i][j]=dp[i-1][j]

如果S[i]=T[j]，dp[i][j]=dp[i-1][j]+j抽出的情况=dp[i-1][j]+dp[i-1][j-1] (注意：这里并不是简单的dp[i-1][j]+1, j抽出后，dp[i-1][j-1]是要大于dp[i-1][j]的)

class Solution {

public:

    int numDistinct(string s, string t) {

        int sLen = s.length();

        int tLen = t.length();

        vector<vector<int>> dp(sLen+, vector<int>(tLen+,));

        for(int i = ; i <= sLen; i++){ //if t==NULL, 1 method to match

            dp[i][]=;

        }

        for(int i = ; i <=sLen; i++){

            for(int j = ; j <= tLen; j++){

                if(s[i-]==t[j-]){

                    dp[i][j]=dp[i-][j]+dp[i-][j-];

                }

                else{

                    dp[i][j]=dp[i-][j];

                }

            }

        }

        return dp[sLen][tLen];

    }

};