代码随想录算法训练营第55天 | 392. 判断子序列、115. 不同的子序列

392. 判断子序列

class Solution {
public:
    bool isSubsequence(string s, string t) {
        if (s.size() > t.size()) return false;
        int i = 0, j = 0;
        while (i < s.size() && j < t.size()) {
            if (s[i] == t[j]) i++;
            j++;
        }
        return i == s.size();
    }
};

动态规划

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int n = s.size(), m = t.size();
        vector> dp(n + 1, vector(m + 1, 0));
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; 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[n][m] == n;
    }
};

115. 不同的子序列

class Solution {
public:
    int numDistinct(string s, string t) {
        int n = s.size(), m = t.size();
        vector> dp(n + 1, vector(m + 1, 0));
        for (int i = 0; i <= n; i++) dp[i][0] = 1;
        for (int j = 1; j <= m; j++) dp[0][j] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; 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[n][m];
    }
};