[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 was formed from the original string by deleting some (can be none) of the C Haracters without disturbing the relative positions of the remaining characters. (ie, is a subsequence of and is not "ACE" "ABCDE" "AEC" ).

Here are an example:
S = "rabbbit" , T ="rabbit"

Return 3 .

Hide TagsDynamic Programming StringMethod One: Dfs, string by case comparison, big data timeout,

classSolution {intm_cnt; Public:    voidDfsConst string& S,Const string& T,intIDX1,intidx2) {           if(IDX2 = =t.size ()) {m_cnt++; return; }           if(Idx1 = =s.size ())return;  for(inti = idx1; I < s.size (); i++)        {               if(S[i] = =t[idx2]) Dfs (s, T, I+1, idx2+1); }       }       intNumdistinct (stringSstringT) {m_cnt=0; DFS (S,t,0,0); returnm_cnt; }   };

Method Two: DP

Set the status to F (i, j), indicating the number of occurrences of t[0,i-1] in s[0,j-1].

If T[i]! = S[j], then f[i][j] = f[i][j-1] means the same number of occurrences as t[0,i-1] in s[0,j-1].

If t[i] = = S[j], then f[i][j] = f[i][j-1] + f[i-1][j-1]

Two methods: T[i] is derived from s[j], there are f[i-1][j-1] species

T[i] is not derived from s[j], but from s[0~i-1], there are f[i][j-1] species

Time complexity O (m*n), Space O (m*n), but can be optimized to O (m) using a scrolling array

classSolution { Public:    intNumdistinct (stringSstringt) {vector<int> tmp (s.size () +1,0); Vector<vector<int> > F (t.size () +1, TMP); //F[i][j] indicates nums that t[0,i-1] is converted by s[0,j-1]f[0][0] =1;//null to NULL with only one         for(inti =1; I < s.size (); i++)//There is the only one-on-one-one-out-of- string changed to nullf[0][i] =1;  for(inti =1; I < t.size (); i++)//There is no -it-that null-to-any stringf[i][0] =0;  for(intj =1; J <= S.size (); J + +)        {             for(inti =1; I <= t.size (); i++)            {                if(I >j) F[i][j]=0; Else                {                    if(t[i-1] = = s[j-1]) F[i][j]= f[i-1][j-1] + f[i][j-1]; ElseF[i][j]= f[i][j-1]; }            }        }        returnf[t.size ()][s.size ()];       }}; 

[Leetcode] Distinct subsequences