[LeetCode OJ] 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.

方法一：用回溯法实现，时间复杂度很高，空间复杂度低，对于小数据可以通过，对大数据会出现Time Limit Exceeded

 int num=;
 void countnum(string S, string T) {
     if(T.size()==)
     {
         num++;
         return;
      }
 
      for(int i=; i<S.size(); i++)
      {
          if(S[i]==T[])
          {
              string s2 = S.substr(i+);
              string t2 = T.substr();
               countnum(s2, t2);
           }
 
      }
      return;
 }
 
 class Solution {
 public:
     int numDistinct(string S, string T) {
         countnum(S, T);
         return num;
     }
 };

方法二：用动态规划（DP）实现，需要的空间复杂度为O(N*M)，对于大数据也可以很快处理。

 class Solution {
 public:
     int numDistinct(string S, string T) {
         vector<vector<int> > num(S.size()+,vector<int>(T.size()+,)); //num[i][j]表示T中的前j个字符构成的子字符串在S中的前i个字符中出现的次数,num[i][j]满足:
         S = " "+ S;                                                                                 //（1）若S[i]=T[j]，则num[i][j] = num[i-1][j]+num[i-1][j-1]；
         T = " "+ T;                                                                                 //（2）若S[i]!=T[j]，则num[i][j] = num[i-1][j]；
         num[][]=;                                                                                //（3）若j>i，则num[i][j]=0。
         for(int i=; i<S.size(); i++)
             for(int j=; j<T.size(); j++)
             {
                 if(j>i)
                 {
                     num[i][j]=;
                     break;
                 }
                 if(S[i]==T[j])
                     num[i][j] = num[i-][j] + num[i-][j-];
                 else
                     num[i][j] = num[i-][j];
             }
             return num[S.size()-][T.size()-];
 
     }
 };