[Leetcode] distinct subsequences

Distinct subsequences

Given a stringSAnd a stringT, Count the number of distinct subsequencesTInS.

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"ABCDE"While"AEC"Is not ).

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

Return3.

 

I wrote a deep-first recursive program, TLE. The leetcode requirement is really high ..

The space is changed to time, and DP is used for this. That is to say, a recursive condition is used to obtain the relationship between N-1 and n steps.

The recursive condition is set to the first I characters of S. The conversion method of the first J characters of T is obtained by deleting the characters. The two-dimensional array element transarr [I] [J] is used for recording.

(1) If s [I] = T [J], add s [I] and T [J] To s [0 ~ I-1] and T [0 ~ The J-1] can fully replicate the transarr [I-1] [J-1] type conversion method. In addition, s [0 ~ I-1] itself (without adding s [I]) can also be converted to T [0 ~ I-1] [J] Mode ~ J].

Transarr [I] [J] = transarr [I-1] [J-1] + transarr [I-1] [J];

(2) If s [I]! = T [J], meaning s [I] cannot be used for conversion to T [0 ~ J], t [J] needs to be from S [0 ~ I-1], so s [0 ~ I] and S [0 ~ I-1.

Transarr [I] [J] = transarr [I-1] [J];

 

However, the above conditions are not enough. The result of transarr [m] [N] is the accumulation of 0. Therefore, you need to define the initial value.

If transarray [I] [0] is set to 1, it means that if it is converted to an empty string of any length, it means to delete all characters in 1.

 

class Solution {public:    int numDistinct(string S, string T)     {        int m = S.size();        int n = T.size();        vector<vector<int>> transArr(m+1, vector<int>(n+1, 0));        for(int i = 0; i < m+1; i ++)            transArr[i][0] = 1;        for(int i = 1; i < m+1; i ++)        {            for(int j = 1; j < n+1; j ++)            {                if(S[i-1] == T[j-1])                    transArr[i][j] = transArr[i-1][j-1]+transArr[i-1][j];                else                    transArr[i][j] = transArr[i-1][j];            }        }        return transArr[m][n];    }};