一刷820

Author: diguage

README.adoc

@@ -5765,12 +5765,12 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 |Easy
 |
 
-//|{counter:codes}
-//|{leetcode_base_url}/short-encoding-of-words/[820. Short Encoding of Words^]
-//|{source_base_url}/_0820_ShortEncodingOfWords.java[Java]
-//|{doc_base_url}/0820-short-encoding-of-words.adoc[题解]
-//|Medium
-//|
+|{counter:codes}
+|{leetcode_base_url}/short-encoding-of-words/[820. Short Encoding of Words^]
+|{source_base_url}/_0820_ShortEncodingOfWords.java[Java]
+|{doc_base_url}/0820-short-encoding-of-words.adoc[题解]
+|Medium
+|
 
 |{counter:codes}
 |{leetcode_base_url}/shortest-distance-to-a-character/[821. Shortest Distance to a Character^]

docs/0820-short-encoding-of-words.adoc

@@ -1,40 +1,78 @@
 [#0820-short-encoding-of-words]
-= 820. Short Encoding of Words
+= 820. 单词的压缩编码
 
-{leetcode}/problems/short-encoding-of-words/[LeetCode - Short Encoding of Words^]
+https://leetcode.cn/problems/short-encoding-of-words/[LeetCode - 820. 单词的压缩编码^]
 
-Given a list of words, we may encode it by writing a reference string `S` and a list of indexes `A`.
+单词数组 `words` 的 *有效编码* 由任意助记字符串 `s` 和下标数组 `indices` 组成，且满足：
 
-For example, if the list of words is `["time", "me", "bell"]`, we can write it as `S = "time#bell#"` and `indexes = [0, 2, 5]`.
+* `words.length == indices.length`
+* 助记字符串 `s` 以 `#` 字符结尾
+* 对于每个下标 `indices[i]` ，`s` 的一个从 `indices[i]` 开始、到下一个 `#` 字符结束（但不包括 `#`）的 *子字符串* 恰好与 `words[i]` 相等
 
-Then for each index, we will recover the word by reading from the reference string from that index until we reach a `"#"` character.
+给你一个单词数组 `words` ，返回成功对 `words` 进行编码的最小助记字符串 `s` 的长度 。
 
-What is the length of the shortest reference string S possible that encodes the given words?
+*示例 1：*
 
-*Example:*
+....
+输入：words = ["time", "me", "bell"]
+输出：10
+解释：一组有效编码为 s = "time#bell#" 和 indices = [0, 2, 5] 。
+words[0] = "time" ，s 开始于 indices[0] = 0 到下一个 '#' 结束的子字符串，如加粗部分所示 "time#bell#"
+words[1] = "me" ，s 开始于 indices[1] = 2 到下一个 '#' 结束的子字符串，如加粗部分所示 "time#bell#"
+words[2] = "bell" ，s 开始于 indices[2] = 5 到下一个 '#' 结束的子字符串，如加粗部分所示 "time#bell#"
+....
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* words = `["time", "me", "bell"]`
-*Output:* 10
-*Explanation:* S = `"time#bell#" and indexes = [0, 2, 5`].
-----
+*示例 2：*
+
+....
+输入：words = ["t"]
+输出：2
+解释：一组有效编码为 s = "t#" 和 indices = [0] 。
+....
+
+*提示：*
 
- 
+* `1 \<= words.length \<= 2000`
+* `1 \<= words[i].length \<= 7`
+* `words[i]` 仅由小写字母组成
 
-*Note:*
 
 
-. `1 <= words.length <= 2000`.
-. `1 <= words[i].length <= 7`.
-. Each word has only lowercase letters.
+== 思路分析
 
+投机取巧的办法：给单词数组按照长度从大到小排列（长单词不可能为短单词的后缀），然后逐个判断已有助记词是否保护单词再决定是否添加。
 
+image::images/0820-10.gif[{image_attr}]
 
+更好的解法是：将单词反转，将单词添加到前缀树中，最后根据前缀树叶子简单对应的长度来计算结果。
+
+image::images/0820-11.jpg[{image_attr}]
 
 [[src-0820]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0820_ShortEncodingOfWords.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0820_ShortEncodingOfWords_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
 
+. https://leetcode.cn/problems/short-encoding-of-words/solutions/173709/dan-ci-de-ya-suo-bian-ma-by-leetcode-solution/[820. 单词的压缩编码 - 官方题解^]
+. https://leetcode.cn/problems/short-encoding-of-words/solutions/174404/99-java-trie-tu-xie-gong-lue-bao-jiao-bao-hui-by-s/[820. 单词的压缩编码 - 99% Trie 吐血攻略，包教包会^]
+. https://leetcode.cn/problems/short-encoding-of-words/solutions/174362/wu-xu-zi-dian-shu-qing-qing-yi-fan-zhuan-jie-guo-j/[820. 单词的压缩编码 - 无需字典树，轻轻一反转，结果就出来^]

docs/images/0820-10.gif

@@ -1732,7 +1732,7 @@ include::0817-linked-list-components.adoc[leveloffset=+1]
 
 include::0819-most-common-word.adoc[leveloffset=+1]
 
-// include::0820-short-encoding-of-words.adoc[leveloffset=+1]
+include::0820-short-encoding-of-words.adoc[leveloffset=+1]
 
 include::0821-shortest-distance-to-a-character.adoc[leveloffset=+1]
 

docs/images/0820-11.jpg

@@ -1016,6 +1016,11 @@ endif::[]
 |{doc_base_url}/0821-shortest-distance-to-a-character.adoc[题解]
 |❌ 双指针。分别从前向后和从后向前遍历。
 
+|{counter:codes}
+|{leetcode_base_url}/short-encoding-of-words/[820. 单词的压缩编码^]
+|{doc_base_url}/0820-short-encoding-of-words.adoc[题解]
+|✅ 排序。按长度倒序，后根据判断是否存在在决定是否插入。更优解是前缀树。
+
 
 |===
 

docs/index.adoc

@@ -0,0 +1,34 @@
+package com.diguage.algo.leetcode;
+
+import java.util.Arrays;
+import java.util.Comparator;
+
+public class _0820_ShortEncodingOfWords {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2026-07-14 22:10:14
+   */
+  public int minimumLengthEncoding(String[] words) {
+    // 按照长度从大到小排列，长度相同则按自然顺序排列
+    // 更长的单词不可能为别的单词的后缀
+    Arrays.sort(words, Comparator.comparing(String::length, Comparator.reverseOrder())
+      .thenComparing(Comparator.naturalOrder()));
+    StringBuilder sb = new StringBuilder();
+    for (String word : words) {
+      if (sb.indexOf(word + "#") >= 0) {
+        continue;
+      }
+      sb.append(word).append('#');
+    }
+    return sb.length();
+  }
+
+  // end::answer[]
+  static void main() {
+    new _0820_ShortEncodingOfWords()
+      .minimumLengthEncoding(new String[]{"feipyxx", "e"});
+//      .minimumLengthEncoding(new String[]{"ctxdic","c"});
+//      .minimumLengthEncoding(new String[]{"time", "me", "bell"});
+  }
+}