Solution: when see question about two strings , DP should be considered first.

We can abstract this question to calculate appear times for string T with length i in string S with length j, which can be represented by numbers[i][j], then through observation and thinking , we can know for numbers[i][j] it should at least equal the numbers[i][j-1] and if T.charAt(i)==S.charAt(j) , numbers[i][j] should also be add numbers[i-1][j-1]

 

 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+));//dp[i][j]表示对应S前i个和T前j个字符的子问题。

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

     }

 };

 int main()

 {

     Solution s;

     string strS("b");

     string strT("a");

     cout << s.numDistinct(strS, strT) << endl;

     return ;

 }