## 一、题目

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`.

## 二、解题思路

• 如果S[i] = T[j], `dp[i][j] = dp[i-1][j-1]+dp[i-1`][j]

• 如果S[i] 不等于 T[j], `dp[i][j] = dp[i-1][j]`

• 初始条件：当T为空字符串时，从任意的S删除几个字符得到T的方法为1

• `dp[0][0] = 1`; // T和S都是空串.

`dp[1 ... S.length() - 1][0] = 1;` // T是空串，S只有一种子序列匹配。

`dp[0][1 ... T.length() - 1] = 0;` // S是空串，T不是空串，S没有子序列匹配。

## 三、解题代码

``````public int numDistincts(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()];
}
``````