# [Leetcode] Distinct subsequences

Source: Internet
Author: User

Problem Description:

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` .

Basic idea:

Dynamic planning methods. To find a recursive type:

F (s,t) = f (s-1,t-1) +f (s-1,t) (when the last character of S is the same as the last character of T)

F (s,t) = f (s-1,t) (when the last character of S is different from the last character of T)

Code:

` public int numdistinct (string S, String T) {//java if (s.length () < T.length ()) return 0;        int sizeS = S.length ();        int sizet = T.length ();            if (SizeS = = Sizet) {if (S.equals (T)) return 1;        else return 0;                } if (Sizet = = 0) return 1;                int [] [] array = new Int[sizes+1][sizet];                for (int i = 0; i < Sizet; i++) array[0][i] = 0;        Char fch = t.charat (0); for (int i = 1; I <=sizeS; i++) {if (fch = = S.charat (i-1)) array[i][0] = array[i-1][0]+            1;        else array[i][0] = array[i-1][0];            } for (int i = 1; I <=sizeS; i++) {Char sch = S.charat (i-1);                                for (int j = 1; J <sizeT; J + +) {char tch = T.charat (j); if (tch = = Sch) Array[i][j] = Array[i-1][j-1]+array[i-1][J];            else array[i][j] = array[i-1][j];    }} return array[sizes][sizet-1]; }`

[Leetcode] Distinct subsequences

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