一刷1839

Author: diguage

README.adoc

@@ -12899,12 +12899,12 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|Medium
 //|
 
-//|{counter:codes}
-//|{leetcode_base_url}/longest-substring-of-all-vowels-in-order/[1839. Longest Substring Of All Vowels in Order^]
-//|{source_base_url}/_1839_LongestSubstringOfAllVowelsInOrder.java[Java]
-//|{doc_base_url}/1839-longest-substring-of-all-vowels-in-order.adoc[题解]
-//|Medium
-//|
+|{counter:codes}
+|{leetcode_base_url}/longest-substring-of-all-vowels-in-order/[1839. Longest Substring Of All Vowels in Order^]
+|{source_base_url}/_1839_LongestSubstringOfAllVowelsInOrder.java[Java]
+|{doc_base_url}/1839-longest-substring-of-all-vowels-in-order.adoc[题解]
+|Medium
+|
 
 //|{counter:codes}
 //|{leetcode_base_url}/maximum-building-height/[1840. Maximum Building Height^]

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

@@ -1,63 +1,57 @@
 [#1839-longest-substring-of-all-vowels-in-order]
-= 1839. Longest Substring Of All Vowels in Order
+= 1839. 所有元音按顺序排布的最长子字符串
 
-{leetcode}/problems/longest-substring-of-all-vowels-in-order/[LeetCode - 1839. Longest Substring Of All Vowels in Order ^]
+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 ^]
 
-A string is considered *beautiful* if it satisfies the following conditions:
+当一个字符串满足如下条件时，我们称它是 *美丽的* ：
 
+* 所有 5 个英文元音字母（`a` ，`e` ，`i` ，`o` ，`u`）都必须 *至少* 出现一次。
+* 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）
 
-* Each of the 5 English vowels (`'a'`, `'e'`, `'i'`, `'o'`, `'u'`) must appear *at least once* in it.
-* The letters must be sorted in *alphabetical order* (i.e. all `'a'`s before `'e'`s, all `'e'`s before `'i'`s, etc.).
+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* 。
 
+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` 。
 
-For example, strings `"aeiou"` and `"aaaaaaeiiiioou"` are considered *beautiful*, but `"uaeio"`, `"aeoiu"`, and `"aaaeeeooo"` are *not beautiful*.
+*子字符串* 是字符串中一个连续的字符序列。
 
-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`.
+*示例 1：*
 
-A *substring* is a contiguous sequence of characters in a string.
+....
+输入：word = "aeiaaioaaaaeiiiiouuuooaauuaeiu"
+输出：13
+解释：最长子字符串是 "aaaaeiiiiouuu" ，长度为 13 。
+....
 
- 
-*Example 1:*
+*示例 2：*
 
-[subs="verbatim,quotes"]
-----
-*Input:* word = "aeiaaio[.underline]#aaaaeiiiiouuu#ooaauuaeiu"
-*Output:* 13
-*Explanation:* The longest beautiful substring in word is "aaaaeiiiiouuu" of length 13.
-----
+....
+输入：word = "aeeeiiiioooauuuaeiou"
+输出：5
+解释：最长子字符串是 "aeiou" ，长度为 5 。
+....
 
-*Example 2:*
+*示例 3：*
 
-[subs="verbatim,quotes"]
-----
-*Input:* word = "aeeeiiiioooauuu[.underline]#aeiou#"
-*Output:* 5
-*Explanation:* The longest beautiful substring in word is "aeiou" of length 5.
+....
+输入：word = "a"
+输出：0
+解释：没有美丽子字符串，所以返回 0 。
+....
 
-----
+*提示：*
 
-*Example 3:*
+* `1 \<= word.length \<= 5 * 10^5^`
+* `word` 只包含字符 `a`，`e`，`i`，`o` 和 `u` 。
 
-[subs="verbatim,quotes"]
-----
-*Input:* word = "a"
-*Output:* 0
-*Explanation:* There is no beautiful substring, so return 0.
-
-----
-
- 
-*Constraints:*
-
-
-* `1 <= word.length <= 5 * 10^5^`
-* `word` consists of characters `'a'`, `'e'`, `'i'`, `'o'`, and `'u'`.
 
 
+== 思路分析
 
+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。
 
-== 思路分析
+image::images/1839-10.png[{image_attr}]
 
+TIP: 看题解，可以只比较字符大小来解答，探索一下解法。
 
 [[src-1839]]
 [tabs]
@@ -84,4 +78,6 @@ include::{sourcedir}/_1839_LongestSubstringOfAllVowelsInOrder.java[tag=answer]
 
 == 参考资料
 
-
+. 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^]
+. 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！
+. 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

@@ -3761,7 +3761,7 @@ include::1823-find-the-winner-of-the-circular-game.adoc[leveloffset=+1]
 
 // include::1838-frequency-of-the-most-frequent-element.adoc[leveloffset=+1]
 
-// include::1839-longest-substring-of-all-vowels-in-order.adoc[leveloffset=+1]
+include::1839-longest-substring-of-all-vowels-in-order.adoc[leveloffset=+1]
 
 // include::1840-maximum-building-height.adoc[leveloffset=+1]
 

docs/index.adoc

@@ -678,6 +678,11 @@ endif::[]
 |{doc_base_url}/0409-longest-palindrome.adoc[题解]
 |✅ 回文
 
+|{counter:codes2503}
+|{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^]
+|{doc_base_url}/1839-longest-substring-of-all-vowels-in-order.adoc[题解]
+|✅ 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。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

logbook/202503.adoc

@@ -0,0 +1,57 @@
+package com.diguage.algo.leetcode;
+
+public class _1839_LongestSubstringOfAllVowelsInOrder {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-05-13 21:48:41
+   */
+  public int longestBeautifulSubstring(String word) {
+    int result = 0;
+    int cnt = 0;
+    char[] chars = new char[]{'a', 'e', 'i', 'o', 'u'};
+    int idx = -1;
+    for (int i = 0; i < word.length(); i++) {
+      char c = word.charAt(i);
+      if (i == 0) {
+        if (c == 'a') {
+          cnt = 1;
+          idx = 0;
+        }
+      } else {
+        if (word.charAt(i - 1) > c) {
+          if (c == 'a') {
+            cnt = 1;
+            idx = 0;
+          } else {
+            cnt = 0;
+            idx = -1;
+          }
+        } else {
+          if (idx >= 0) {
+            if (c == chars[idx]) {
+              cnt++;
+            } else if (idx < chars.length - 1 && c == chars[idx + 1]) {
+              idx++;
+              cnt++;
+            } else {
+              cnt = 0;
+              idx = -1;
+            }
+          }
+        }
+      }
+      if (idx == chars.length - 1) {
+        result = Math.max(result, cnt);
+      }
+    }
+    return result;
+  }
+
+  // end::answer[]
+  public static void main(String[] args) {
+    new _1839_LongestSubstringOfAllVowelsInOrder()
+      .longestBeautifulSubstring("aeiaaioaaaaeiiiiouuuooaauuaeiu");
+  }
+}