Given a stringSand a stringT, count the number of distinct subsequences ofSwhich equalsT.

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"

Return3.

Solution: dp[i][j]: S前i个字母和T前j个字母有多少个配对subsequence。看S第i个字母是否与T第j个字母是否配对

dp[i][j] = dp[i - 1][j] + (dp[i - 1][j - 1] if s[i - 1] == t[j - 1])

初始条件：

– 如果B是空串，B在A中出现次数是1

– f[i][0] = 1 (i = 0, 1, 2, …, m)

– 如果A是空串而B不是空串，B在A中出现次数是0

– f[0][j] = 0 (j = 1, 2, …, n)

    public int numDistinct(String s, String t) {
        if (s == null || s.length() == 0 || t == null || t.length() == 0) {
            return 0;
        }
        int m = s.length(), n = t.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 0; i <= m; i++) {
            dp[i][0] = 1;
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                dp[i][j] = dp[i - 1][j];
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    dp[i][j] += dp[i - 1][j - 1];
                }
            }
        }
        return dp[m][n];
    }