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‎README.adoc

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//|{counter:codes}
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//|{leetcode_base_url}/longest-substring-of-all-vowels-in-order/[1839. Longest Substring Of All Vowels in Order^]
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//|{source_base_url}/_1839_LongestSubstringOfAllVowelsInOrder.java[Java]
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//|{doc_base_url}/1839-longest-substring-of-all-vowels-in-order.adoc[题解]
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//|
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|{counter:codes}
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|{leetcode_base_url}/longest-substring-of-all-vowels-in-order/[1839. Longest Substring Of All Vowels in Order^]
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|{source_base_url}/_1839_LongestSubstringOfAllVowelsInOrder.java[Java]
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|{doc_base_url}/1839-longest-substring-of-all-vowels-in-order.adoc[题解]
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//|{counter:codes}
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//|{leetcode_base_url}/maximum-building-height/[1840. Maximum Building Height^]

‎docs/1839-longest-substring-of-all-vowels-in-order.adoc

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[#1839-longest-substring-of-all-vowels-in-order]
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= 1839. Longest Substring Of All Vowels in Order
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= 1839. 所有元音按顺序排布的最长子字符串
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{leetcode}/problems/longest-substring-of-all-vowels-in-order/[LeetCode - 1839. Longest Substring Of All Vowels in Order ^]
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https://leetcode.cn/problems/longest-substring-of-all-vowels-in-order/[LeetCode - 1839. 所有元音按顺序排布的最长子字符串 ^]
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A string is considered *beautiful* if it satisfies the following conditions:
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当一个字符串满足如下条件时，我们称它是 *美丽的* ：
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* 所有 5 个英文元音字母（`a` ，`e` ，`i` ，`o` ，`u`）都必须 *至少* 出现一次。
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* 这些元音字母的顺序都必须按照 *字典序* 升序排布（也就是说所有的 `a` 都在 `e` 前面，所有的 `e` 都在 `i` 前面，以此类推）
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* Each of the 5 English vowels (`'a'`, `'e'`, `'i'`, `'o'`, `'u'`) must appear *at least once* in it.
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* The letters must be sorted in *alphabetical order* (i.e. all `'a'`s before `'e'`s, all `'e'`s before `'i'`s, etc.).
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比方说，字符串 `aeiou` 和 `aaaaaaeiiiioou` 都是 *美丽的* ，但是 `uaeio` ，`aeoiu` 和 `aaaeeeooo` *不是美丽的* 。
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给你一个只包含英文元音字母的字符串 `word` ，请你返回 `word` 中 *最长美丽子字符串的长度* 。如果不存在这样的子字符串，请返回 `0` 。
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For example, strings `"aeiou"` and `"aaaaaaeiiiioou"` are considered *beautiful*, but `"uaeio"`, `"aeoiu"`, and `"aaaeeeooo"` are *not beautiful*.
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*子字符串* 是字符串中一个连续的字符序列。
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Given a string `word` consisting of English vowels, return _the *length of the longest beautiful substring* of _`word`_. If no such substring exists, return _`0`.
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*示例 1：*
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A *substring* is a contiguous sequence of characters in a string.
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....
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输入：word = "aeiaaioaaaaeiiiiouuuooaauuaeiu"
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输出：13
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解释：最长子字符串是 "aaaaeiiiiouuu" ，长度为 13 。
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....
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*Example 1:*
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*示例 2：*
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[subs="verbatim,quotes"]
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----
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*Input:* word = "aeiaaio[.underline]#aaaaeiiiiouuu#ooaauuaeiu"
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*Output:* 13
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*Explanation:* The longest beautiful substring in word is "aaaaeiiiiouuu" of length 13.
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----
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....
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输入：word = "aeeeiiiioooauuuaeiou"
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输出：5
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解释：最长子字符串是 "aeiou" ，长度为 5 。
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....
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*Example 2:*
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*示例 3：*
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[subs="verbatim,quotes"]
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----
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*Input:* word = "aeeeiiiioooauuu[.underline]#aeiou#"
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*Output:* 5
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*Explanation:* The longest beautiful substring in word is "aeiou" of length 5.
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....
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输入：word = "a"
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输出：0
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解释：没有美丽子字符串，所以返回 0 。
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....
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----
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*提示：*
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*Example 3:*
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* `1 \<= word.length \<= 5 * 10^5^`
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* `word` 只包含字符 `a`，`e`，`i`，`o` 和 `u` 。
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[subs="verbatim,quotes"]
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----
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*Input:* word = "a"
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*Output:* 0
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*Explanation:* There is no beautiful substring, so return 0.
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----
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*Constraints:*
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* `1 <= word.length <= 5 * 10^5^`
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* `word` consists of characters `'a'`, `'e'`, `'i'`, `'o'`, and `'u'`.
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== 思路分析
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其实就是一个字符判断，当遇到第一个 `a` 时开始计数，符合要求（和上一个字符相同，或者是下一个元音字符）计数加一，否则归零。当到达最后一个元音字符时，开始更新结果值，保留最大值。
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== 思路分析
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image::images/1839-10.png[{image_attr}]
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TIP: 看题解，可以只比较字符大小来解答，探索一下解法。
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[[src-1839]]
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[tabs]
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== 参考资料
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. https://leetcode.cn/problems/longest-substring-of-all-vowels-in-order/solutions/742722/suo-you-yuan-yin-an-shun-xu-pai-bu-de-zu-9wqg/[1839. 所有元音按顺序排布的最长子字符串 - 官方题解^]
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. https://leetcode.cn/problems/longest-substring-of-all-vowels-in-order/solutions/742704/bi-da-xiao-by-sweetpepperj-gdlt/[1839. 所有元音按顺序排布的最长子字符串 - 不用元音判断，直接通过比大小解题^] -- 这个解法非常轻巧！
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. https://leetcode.cn/problems/longest-substring-of-all-vowels-in-order/solutions/3674813/tan-xin-hua-dong-chuang-kou-c-by-lukasir-l7la/[1839. 所有元音按顺序排布的最长子字符串 - 贪心+滑动窗口(C++)^]

‎docs/images/1839-10.png

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‎docs/index.adoc

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‎logbook/202503.adoc

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|{doc_base_url}/0409-longest-palindrome.adoc[题解]
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|✅ 回文
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|{counter:codes2503}
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|{leetcode_base_url}/longest-substring-of-all-vowels-in-order/[1839. 所有元音按顺序排布的最长子字符串^]
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|{doc_base_url}/1839-longest-substring-of-all-vowels-in-order.adoc[题解]
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|✅ 滑动窗口。题目不难，只是处理边界条件麻烦。另外，也有取巧解法，可以探索一下。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_1839_LongestSubstringOfAllVowelsInOrder.java

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package com.diguage.algo.leetcode;
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public class _1839_LongestSubstringOfAllVowelsInOrder {
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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 2025-05-13 21:48:41
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*/
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public int longestBeautifulSubstring(String word) {
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int result = 0;
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int cnt = 0;
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char[] chars = new char[]{'a', 'e', 'i', 'o', 'u'};
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int idx = -1;
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for (int i = 0; i < word.length(); i++) {
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char c = word.charAt(i);
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if (i == 0) {
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if (c == 'a') {
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cnt = 1;
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idx = 0;
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}
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} else {
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if (word.charAt(i - 1) > c) {
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if (c == 'a') {
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cnt = 1;
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idx = 0;
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} else {
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cnt = 0;
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idx = -1;
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}
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} else {
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if (idx >= 0) {
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if (c == chars[idx]) {
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cnt++;
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} else if (idx < chars.length - 1 && c == chars[idx + 1]) {
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idx++;
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cnt++;
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} else {
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cnt = 0;
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idx = -1;
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}
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}
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}
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}
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if (idx == chars.length - 1) {
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result = Math.max(result, cnt);
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}
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}
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return result;
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}
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// end::answer[]
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public static void main(String[] args) {
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new _1839_LongestSubstringOfAllVowelsInOrder()
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.longestBeautifulSubstring("aeiaaioaaaaeiiiiouuuooaauuaeiu");
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}
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}

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