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一刷392
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README.adoc

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@@ -2769,14 +2769,14 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
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//|{doc_base_url}/0391-perfect-rectangle.adoc[题解]
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//|Hard
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//|
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//
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//|{counter:codes}
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//|{leetcode_base_url}/is-subsequence/[392. Is Subsequence^]
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//|{source_base_url}/_0392_IsSubsequence.java[Java]
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//|{doc_base_url}/0392-is-subsequence.adoc[题解]
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//|Easy
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//|
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//
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|{counter:codes}
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|{leetcode_base_url}/is-subsequence/[392. Is Subsequence^]
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|{source_base_url}/_0392_IsSubsequence.java[Java]
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|{doc_base_url}/0392-is-subsequence.adoc[题解]
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|Easy
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|
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//|{counter:codes}
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//|{leetcode_base_url}/utf-8-validation/[393. UTF-8 Validation^]
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//|{source_base_url}/_0393_UTF8Validation.java[Java]

docs/0392-is-subsequence.adoc

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Special thanks to <a href="https://leetcode.com/pbrother/">@pbrother</a> for adding this problem and creating all test cases.
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== 思路分析
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双指针,当短指针到头也就是表明是子串。
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[[src-0392]]
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[tabs]
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====
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一刷::
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+
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0392_IsSubsequence.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0392_IsSubsequence_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/is-subsequence/solutions/1658262/by-jyd-zeph/?envType=study-plan-v2&envId=selected-coding-interview[392. 判断子序列 - 双指针,清晰图解^]
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docs/index.adoc

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// include::0391-perfect-rectangle.adoc[leveloffset=+1]
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// include::0392-is-subsequence.adoc[leveloffset=+1]
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include::0392-is-subsequence.adoc[leveloffset=+1]
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// include::0393-utf-8-validation.adoc[leveloffset=+1]
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logbook/202406.adoc

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|{doc_base_url}/0169-majority-element.adoc[题解]
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|⭕️ 哈希计数法最易想到,摩尔投票法最妙!
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|{counter:codes}
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|{leetcode_base_url}/is-subsequence/[392. Is Subsequence^]
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|{doc_base_url}/0392-is-subsequence.adoc[题解]
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|✅ 双指针
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|===
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截止目前,本轮练习一共完成 {codes} 道题。
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package com.diguage.algo.leetcode;
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public class _0392_IsSubsequence {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2024-09-20 21:50:25
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*/
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public boolean isSubsequence(String s, String t) {
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if (s == null || s.length() == 0) {
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return true;
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}
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int idx = 0;
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for (int i = 0; i < t.length(); i++) {
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if (idx >= s.length()){
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break;
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}
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if (t.charAt(i) == s.charAt(idx)) {
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idx++;
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}
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}
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return idx >= s.length();
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}
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// end::answer[]
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}

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