LeetCode 392 Is Subsequence

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几百个测试用例一致通过

最新推荐文章于 2024-07-10 20:26:14 发布

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分类专栏：算法文章标签：算法 leetcode string

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本文介绍了一个简单的算法来解决子序列检查问题，即判断一个字符串是否为另一个字符串的子序列。通过遍历较短字符串并在较长字符串中查找相应字符，确保了字符间的相对顺序不变。

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Given a string s and a string t, check if s is subsequence of t.
You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (< =100). A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not). Example 1: s = "abc", t = "ahbgdc" Return true. Example 2: s = "axc", t = "ahbgdc" Return false. 翻译：给你两个字符串，判断s是不是t的子串，s的每个字符在t中的相对顺序不能边，但是可以不连续。比如abc是abcde的子串，也是ahbgdc的子串，但是不是ahbgdc的子串。这个题我用了一个简单粗暴的方法，最后还超过了93.52%的提交。算法：假如我们已经确定s的前i位是t的子串，如果s[i+1]在t中出现那么s可能是t的子串，否则一定不是子串。

/**
     * @param s
     * @param t
     * @return
     */
    public static boolean isSubsequence(String s, String t) {
        if (s.length() > t.length())
            return false;
        char[] sArr = s.toCharArray();
        int index = -1;
        for (int i = 0; i < sArr.length; i++) {
            index = t.indexOf(s.charAt(i), index + 1);
            if (index == -1)
                return false;
        }
        return true;
    }