代码随想录算法训练营第54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) {
        int n = s.length(), m = t.length();
        int i = 0, j = 0;
        while (i < n && j < m) {
            if (s[i] == t[j]) {
                i++;
            }
            j++;
        }
        return i == n;
    }
};

java版：

class Solution {
    public boolean isSubsequence(String s, String t) {
        int n = s.length(), m = t.length();
        int i = 0, j = 0;
        while (i < n && j < m) {
            if (s.charAt(i) == t.charAt(j)) {
                i++;
            }
            j++;
        }
        return i == n;
    }
}

115. 不同的子序列

class Solution {
public:
    int numDistinct(string s, string t) {
        int m = s.length(), n = t.length();
        if (m < n) {
            return 0;
        }
        vector> dp(m + 1, vector(n + 1));
        for (int i = 0; i <= m; i++) {
            dp[i][n] = 1;
        }
        for (int i = m - 1; i >= 0; i--) {
            char sChar = s.at(i);
            for (int j = n - 1; j >= 0; j--) {
                char tChar = t.at(j);
                if (sChar == tChar) {
                    dp[i][j] = dp[i + 1][j + 1] + dp[i + 1][j];
                } else {
                    dp[i][j] = dp[i + 1][j];
                }
            }
        }
        return dp[0][0];
    }
};

java版：

class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length(), n = t.length();
        if (m < n) {
            return 0;
        }
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 0; i <= m; i++) {
            dp[i][n] = 1;
        }
        for (int i = m - 1; i >= 0; i--) {
            char sChar = s.charAt(i);
            for (int j = n - 1; j >= 0; j--) {
                char tChar = t.charAt(j);
                if (sChar == tChar) {
                    dp[i][j] = dp[i + 1][j + 1] + dp[i + 1][j];
                } else {
                    dp[i][j] = dp[i + 1][j];
                }
            }
        }
        return dp[0][0];
    }
}