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.

这一题使用DR来做最简洁。 如果用burust的方法也行，但是时间复杂度很高。。。。

n(T.length())

m(S.length())

[m,n] 表示在S的前m个字符里， T的前n个字符组成的字符串的出现次数。

故：[m,n] = [m -1, n] + [m-1,n-1]a,  (a = 1 if(S.charAt(m) == T.charAt(n)).

然后还需要判断一下test case。

public class Solution {
    public int numDistinct(String S, String T) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(T.length() == 0 || S.length() == 0) return 0;
        
        int[][] map = new int[S.length()][T.length()];
        map[0][0] = (S.charAt(0) == T.charAt(0)) ? 1 : 0;
        int count = 0;
        for(int i = 0; i < S.length(); i ++){
            if(S.charAt(i) == T.charAt(0)) count ++;
            map[i][0] = count;
        }
        
        for(int i = 1; i < T.length(); i ++){
            for(int j = 1; j < S.length(); j ++){
                map[j][i] = map[j-1][i];
                if(S.charAt(j) == T.charAt(i)){
                    map[j][i] += map[j-1][i-1];
                }
            }
        }
        return map[S.length()-1][T.length()-1];
    }
}