# 115.不同的子序列

## 问题描述
[115.不同的子序列](https://leetcode.cn/problems/distinct-subsequences/)

## 解题思路
`dp[i][j]`表示考虑考虑`t`的前`j`个字符在`s`的前`i`个字符中的出现个数:
- `if (s[i - 1] == t[j - 1]) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];`(表示使用`s[i - 1]`匹配和不使用`s[i - 1]`匹配)
- `else dp[i][j] = dp[i - 1][j];`

## 代码
```cpp
class Solution {
public:
    int numDistinct(string s, string t) {
        if (s.size() < t.size())
            return 0;
        vector<vector<uint32_t>> dp(s.size() + 1, vector<uint32_t>(t.size() + 1, 0));
        // dp[0][0] = 1;
        for (int i = 0; i <= s.size(); i++) {
            dp[i][0] = 1;
        }
        for (int i = 1; i <= s.size(); i++) {
            for (int j = 1; j <= i && j <= t.size(); j++) {
                if (s[i - 1] == t[j - 1])
                    dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
                else
                    dp[i][j] = dp[i - 1][j];
            }
        }
        return dp[s.size()][t.size()];
        
    }
};
```