392. Is Subsequence(easy)_subsequences easy-优快云博客

本文链接：https://blog.youkuaiyun.com/SourDumplings/article/details/103027497

本文介绍了一种检查字符串s是否为另一字符串t的子序列的算法，通过双指针技术实现，适用于s长度小于等于100，t长度可能高达500,000的情况。文章提供了C++和Java的实现代码，并讨论了当有大量输入S时如何优化代码。

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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.

Follow up:
If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

C++：

/*
 * @Autor: SourDumplings
 * @Date: 2019-09-21 17:09:37
 * @Link: https://github.com/SourDumplings/
 * @Email: changzheng300@foxmail.com
 * @Description: https://leetcode.com/problems/is-subsequence/
 */

class Solution
{
public:
    bool isSubsequence(string s, string t)
    {
        int ls = s.length();
        int lt = t.length();
        int j = 0;
        for (int i = 0; i < ls; i++)
        {
            for (; j < lt; j++)
            {
                if (s[i] == t[j])
                {
                    j++;
                    break;
                }
            }
            if (j == lt && s[i] != t[j - 1])
            {
                return false;
            }
        }
        return true;
    }
};

Java：

/*
 * @Autor: SourDumplings
 * @Date: 2019-09-21 17:42:07
 * @Link: https://github.com/SourDumplings/
 * @Email: changzheng300@foxmail.com
 * @Description: https://leetcode.com/problems/is-subsequence/
 */

class Solution
{
    public boolean isSubsequence(String s, String t)
    {
        int j = 0;
        int ls = s.length();
        int lt = t.length();
        if (lt == 0 && ls > 0)
        {
            return false;
        }
        for (int i = 0; i < ls; i++)
        {
            char ci = s.charAt(i);
            for (; j < lt; j++)
            {
                char cj = t.charAt(j);
                if (ci == cj)
                {
                    ++j;
                    break;
                }
            }
            if (j == lt && ci != t.charAt(j - 1))
            {
                return false;
            }
        }
        return true;
    }
}