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.

　　

class Solution {
public:

    void DFS(string S, string T,int si,int num, vector<char> &tp)
    {

      if(num == sizeB){
             answer ++;
             return ;
      }

      if(si >= sizeA || num > sizeB)
         return ;

      for(int i = si; i<sizeA ; i++)
      {
          if(S[i] == T[num])
            {
         tp.push_back(S[i]) ;
         DFS(S,T,i+, num+, tp);
         tp.pop_back() ;

         }
      }
    }
    int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sizeA = S.size();
        sizeB = T.size();

        if(sizeA < sizeB) return ;

         answer = ;
         int i;
         for( i= ; i< sizeA  ; i++)
               if(S[i] == T[])
                  break ;

         if(i < sizeA)
         {

           vector<char> tp;
           DFS(S,T,i,,tp) ;

          }

          return answer;
    }
private :
  int answer ;
  int sizeA;
  int sizeB;

};

上述代码使用DFS来做，大数据还过不了

DP:

将“S串中前m个字母的子串中包含多少个T串中前n个字母的子串”这样的问题记为A[m][n]。 可以得到递推式 ：
if(S[m-1] == T[n-1]) A[m][n] = A[m-1][n-1] + A[m-1][n];
else A[m][n] = A[m-1][n-1];
再处理边界情况即可。简单起见，用类似打表记录式的递归实现。

class Solution {
public:

    int Dp(int m, int n, int *tp, const string &S,const string & T )
    {

       if(n == -) return ;
        else if(m == -) return ;

        if(m < n) return ;
       if( tp[m*sizeB+n] != - )
         return tp[m*sizeB+n];

        tp[m*sizeB+n] =  S[m] == T[n] ? Dp(m-, n-,tp, S, T) + Dp(m-, n,tp,S,T) :
                            Dp(m-, n,tp,S,T) ;

        return tp[m*sizeB+n];
    }

    int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sizeA = S.size();
        sizeB = T.size();
       int *tp = new int[sizeA * sizeB] ;
       for(int i = ; i< sizeA * sizeB ;i++)
           tp[i] = -;

       return Dp(sizeA-, sizeB-,tp,S, T) ;

    }

 private :

    int sizeA;
    int sizeB;
};

上面必须先判断n == -1,在判断m == -1. 很重要。

一种写法：

class Solution {
public:
    int numDistinct(string S, string T) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
     //if(S == null || T == null) return -1;
     int lens = S.size();
     int lent = T.size();

     if(lens ==  || lent == ) return ;
     vector<vector<int>> m(lent+,vector<int>(lens+,));
     for(int j = ; j <= S.length(); j++) m[][j] = ;

     for(int i = ; i <= lent; i++)
        for(int j = i; j <= lens; j++)
            m[i][j] = T[i-] != S[j-] ? m[i][j-] : m[i-][j-] + m[i][j-];

     return m[lent][lens];
    }
};

节省空间的写法：

class Solution {
public:
    int numDistinct(string S, string T) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
    int M = T.length(); //subsequence length
    int N = S.length();
    if (M > N || M == || N==) {
        return ;
    }
    vector<int> m(M, );
    m[] = (T[] == S[]?:);
    for (int i=; i<N; ++i) {
        for(int j=min(i,M); j>=;--j) {
            m[j] = m[j] + ((S[i]==T[j])?m[j-]:);
        }
        m[] = m[] + (S[i]==T[]?:);
    }
    return m[M-];
    }
};