[Leetcode] Thoughts on solving 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 .

Question: Given the string s, T, how many different sub-sequences in s are available for T.

The first to see this problem, no ideas, on the Internet to see other people's answers to understand how to solve problems. This is a DP problem, if you want to get the state escape relationship, the answer is more obvious. The solution is a bit like knapsack problem solution.

A two-dimensional array Vv[t.len][s.len] stores intermediate results. Vv[i][k], which represents the number of different sub-sequences in t[0, I] in S[0, K].

State transition Relationship, text description:

When S[i] and t[k] are unequal, the number of different sub-sequences of t[0, I] in S[0, K], and the number of different sub-sequences in S[0, k-1] are the same.

When S[i] and T[k] are at the same time, t[0, I] in S[0, K] The number of different sub-sequences equal to S[i] is taken (vv[i-1][k-1]), plus s[i] is not taken (vv[i][k-1]).

State transition relationships, which are represented by a formula:

When s[i]! = T[k], vv[i][k] = vv[i][k-1]

When s[i] = = T[k], vv[i][k] = Vv[i-1][k-1] + vv[i][k-1]

1 intNumdistinct (stringSstringt) {2     3     if(T.size () >s.size ()) {4         return 0;5     }6 7vector<vector<int>> VV (T.size (), vector<int> (S.size (),-1));8     9     //Boundary Value ProcessingTen      for(inti =1; I < vv.size (); i++) { Onevv[i][i-1] =0; A     } -  -     //Boundary Value Processing the     if(s[0] = = t[0]) { -vv[0][0] =1; -}Else{ -vv[0][0] =0; +     } -      +     //Boundary Value Processing A      for(inti =1; I < vv[0].size (); i++) { at         if(t[0] ==S[i]) { -vv[0][i] = vv[0][i-1] +1; -}Else{ -vv[0][i] = vv[0][i-1]; -         } -     } in      -     //State Transitions to      for(inti =1; I < vv.size (); i++) { +          for(intK = i; K < Vv[i].size (); k++) { -             if(T[i]! =S[k]) { theVv[i][k] = vv[i][k-1]; *}Else{ $Vv[i][k] = vv[i-1][k-1] + vv[i][k-1];Panax Notoginseng             } -         } the     } +      A     returnVv[t.size ()-1][s.size ()-1]; the}

[Leetcode] Thoughts on solving Distinct subsequences