392. Is Subsequence

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DP

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class Solution {
    public boolean isSubsequence(String s, String t) {
        int m = t.length(), n = s.length();

        // 定义 dp[i][j] 为字符串 t 前 i 个字符是否存在子序列 s[0, j - 1]
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;
        for (int i = 1; i <= m; i++) {
            dp[i][0] = true;
        }
        for (int j = 1; j <= n; j++) {
            dp[0][j] = false;
        }

        for (int i = 1; i <= m; i++) {
            char a = t.charAt(i - 1);
            for (int j = 1; j <= n; j++) {
                char b = s.charAt(j - 1);

                // 目标：b 一定要被匹配
                if (a == b) {
                    // 使用 a 去匹配 b
                    dp[i][j] = dp[i - 1][j - 1];
                    // 不使用 a 去匹配 b
                    dp[i][j] |= dp[i - 1][j];
                } else {
                    // 无法使用 a 去匹配 b
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }

        return dp[m][n];
    }
}

Two Pointers

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class Solution {
    public boolean isSubsequence(String s, String t) {
        int m = s.length(), n = t.length();

        int i = 0;
        for (int j = 0; i < m && j < n; j++) {
            if (s.charAt(i) == t.charAt(j)) {
                i++;
            }
        }

        return i == m;
    }
}

References

392. Is Subsequence