distinct sequence Google

leetcode ,暴力解法dfs, 双指针，不行

public int numDistinct(String s, String t) {
return numDistinctHelper(s,t,0,0);
}
public int numDistinctHelper(String s, String t, int si, int ti){
if(ti == t.length()) {
return 1;
}
if(si >= s.length()){
return 0;
}
int count = 0;
for(int i = si; i < s.length(); i++){
if(s.charAt(i) == t.charAt(ti)){
count += numDistinctHelper(s,  t, i+1, ti+1);
}
}
return count;
}

A Clever way to solve it: (Dynamic Programming)

solution: Let W(i, j) stand for the number of subsequences of S(0, i) equals to T(0, j). If S.charAt(i) == T.charAt(j), W(i, j) = W(i-1, j-1) + W(i-1,j); Otherwise, W(i, j) = W(i-1,j).

public int numDistinct(String S, String T) {
int[][] table = new int[S.length() + 1][T.length() + 1];

for (int i = 0; i < S.length(); i++)
table[i][0] = 1;

for (int i = 1; i <= S.length(); i++) {
for (int j = 1; j <= T.length(); j++) {
if (S.charAt(i - 1) == T.charAt(j - 1)) {
table[i][j] += table[i - 1][j] + table[i - 1][j - 1];
} else {
table[i][j] += table[i - 1][j];
}
}
}

return table[S.length()][T.length()];
}