算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Distinct Subsequences


### 描述

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.


### 分析

设状态为`f(i,j)`,表示`T[0,j]`在`S[0,i]`里出现的次数。首先,无论`S[i]`和`T[j]`是否相等,若不使用`S[i]`,则`f(i,j)=f(i-1,j)`;若`S[i]==T[j]`,则可以使用`S[i]`,此时`f(i,j)=f(i-1,j)+f(i-1, j-1)`。


### 代码

{% if book.java %}
{% codesnippet "./code/distinct-subsequences."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
```cpp
// Distinct Subsequences
// 二维动规+滚动数组
// 时间复杂度O(m*n),空间复杂度O(n)
class Solution {
public:
int numDistinct(const string &S, const string &T) {
vector f(T.size() + 1);
f[0] = 1;
for (int i = 0; i < S.size(); ++i) {
for (int j = T.size() - 1; j >= 0; --j) {
f[j + 1] += S[i] == T[j] ? f[j] : 0;
}
}

return f[T.size()];
}
};
```
{% endif %}