Given a string S and a string T, count the number of distinct subsequences of S which equals T.

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).

Example 1:
Input: S = "rabbbit", T = "rabbit"
Output: 3
Explanation:
As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)
rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^
Example 2:
Input: S = "babgbag", T = "bag"
Output: 5
Explanation:
As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)
babgbag
^^ ^
babgbag
^^    ^
babgbag
^    ^^
babgbag
  ^  ^^
babgbag
    ^^^

dp[i][j]代表s的前i个字符挑出t的前j个字符的方案个数
class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length();
        int n = t.length();
        if(n > m) return 0;

        int[][] dp = new int[m+1][n+1];
        for(int i = 0; i < m+1; i++) {
            dp[i][0] = 1;
        }
        for(int i = 1; i < n+1; i++) {
            dp[0][i] = 0;
        }

        for(int i = 1; i < m+1; i++) {
            for(int j = 1; j < n+1; j++) {
                if(s.charAt(i-1) == t.charAt(j-1)) dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
                //                                           第i个和第j个配上  不配上，那就看前i-1个和前j个 （这是两种不同的方案 要加起来）
                else dp[i][j] = dp[i-1][j];
            }
        }
        return dp[m][n];
    }
}