还是比较头疼这种两个字符串，用二维下标表示状态的题目。

392. 判断子序列

也可以双指针

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int n1 = s.size(), n2 = t.size();
        vector<vector<int>> dp(n1 + 1, vector<int> (n2 + 1));
        for (int i = 1; i <= n1; i ++)
        {
            for (int j = 1; j <= n2; j ++)
            {
                if (s[i-1] == t[j-1]) dp[i][j] = dp[i-1][j-1] + 1;
                else dp[i][j] = dp[i][j-1];
            }
        }
        return dp[n1][n2] == s.size();
    }
};

115. 不同的子序列

class Solution {
public:
    // 相同时可以选或不选当前下标
    int numDistinct(string s, string t) {
        int n1 = s.size(), n2 = t.size();
        vector<vector<unsigned long long>> dp(n1 + 1, vector<unsigned long long> (n2 + 1));
        for (int i = 0; i <= n1; i ++) dp[i][0] = 1;
        for (int i = 1; i <= n1; i ++)
        {
            for (int j = 1; j <= n2; j ++)
            {
                if (s[i-1] == t[j-1])
                {
                    dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
                }
                else
                    dp[i][j] = dp[i-1][j];
            }
        }
        return dp[n1][n2];
    }
};