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

392.判断子序列

讲解链接：代码随想录-392.判断子序列

动态规划

public boolean isSubsequence(String s, String t) {
    int[][] dp = new int[s.length() + 1][t.length() + 1];
    for (int i = 1; i <= s.length(); i++) {
        for (int j = 1; j <= t.length(); j++) {
            if (s.charAt(i - 1) == t.charAt(j - 1)) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            } else {
                dp[i][j] = dp[i][j - 1];
            }
        }
    }
  
    if (dp[s.length()][t.length()] == s.length()) {
        return true;
    } else {
        return false;
    }
}

双指针法

public boolean isSubsequence(String s, String t) {
    int left = 0;
    for (int i = 0; i < t.length() && left < s.length(); i++) {
        if (s.charAt(left) == t.charAt(i)) {
            left++;
        }
    }
    return left == s.length();
}

115.不同的子序列

讲解链接：代码随想录-115.不同的子序列

public int numDistinct(String s, String t) {
    int[][] dp = new int[s.length() + 1][t.length() + 1];
    for (int i = 0; i <= s.length(); i++) {
        dp[i][0] = 1;
    }
    for (int j = 1; j <= t.length(); j++) {
        dp[0][j] = 0;
    }
    for (int i = 1; i <= s.length(); i++) {
        for (int j = 1; j <= t.length(); j++) {
            if (s.charAt(i - 1) == t.charAt(j - 1)) {
                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
            } else {
                dp[i][j] = dp[i - 1][j];
            }
        }
    }

    return dp[s.length()][t.length()];
}