chore(docs): sync doc metadata [skip ci]

Author: longsizhuo

app/docs/CommunityShare/Leetcode/2131. 连接两字母单词得到的最长回文串.md

@@ -1,10 +1,11 @@
 ---
 title: 2131. 连接两字母单词得到的最长回文串.md
-date: '2025/5/25-2:33'
+date: "2025/5/25-2:33"
 tags:
   - - Python
   - - Answer
 abbrlink: 9fa195e5
+docId: ksw2vic4alf1tdnnueay81g8
 ---
 
 # QUESTION：
@@ -34,18 +35,18 @@ Example:
 
 5. Second "gg" finds "gg" exists → use "gg" + "gg" as a pair → res += 4 → map becomes { "lc": 0, "gg": 0 }
 
-    Then we check whether the hash map contains any palindromic word (i.e., a word with two identical characters) that can be used as the center of the final palindrome string.
-    If such a word exists, we can add 2 more to the result.
+   Then we check whether the hash map contains any palindromic word (i.e., a word with two identical characters) that can be used as the center of the final palindrome string.
+   If such a word exists, we can add 2 more to the result.
 
 6. Finding a center word (can only pick one symmetric word)
-⚠️ In this example, all "gg" words have been paired, so none is left → no center word is added.
-
+   ⚠️ In this example, all "gg" words have been paired, so none is left → no center word is added.
 
 思路来自ling-nc大佬. 构建哈希表，统计每个单词的出现次数。
 
 对于每个单词，检查其逆序是否存在于哈希表中，如果存在，则可以形成一个回文串，更新结果并减少对应的计数。
 如果逆序不存在，则将该单词的计数增加。最后检查是否有可以作为中心的回文单词。
 举例:
+
 1. 输入 ["lc", "cl", "gg", "gg"]
 
 2. "lc" 没有配对 → 存入 map: { "lc": 1 }
@@ -55,10 +56,11 @@ Example:
 4. "gg" 没有配对 → 存入 map: { "lc": 0, "gg": 1 }
 
 5. 第二个 "gg" 发现 "gg" 存在 → 成对使用 "gg"+"gg" → res += 4 → map: { "lc": 0, "gg": 0 }
-接着检查哈希表中是否有可以作为中心的回文单词（即两个字母相同的单词），如果有，则可以增加2到结果中。 
+   接着检查哈希表中是否有可以作为中心的回文单词（即两个字母相同的单词），如果有，则可以增加2到结果中。
 
 6. 第二部分：寻找中间部分（只能选一个对称字符串）
-⚠️ 在这个例子中，"gg" 都被配完了，没剩下 → 不会加中间部分。
+   ⚠️ 在这个例子中，"gg" 都被配完了，没剩下 → 不会加中间部分。
+
 # Code：
 
 ```python
@@ -83,28 +85,28 @@ class Solution:
 
 ```typescript
 function longestPalindrome(words: string[]): number {
-    const count: Map<string, number> = new Map();
-    let res = 0;
-
-    for (const word of words) {
-        const reversed = word.split('').reverse().join('');
-        const reversedCount = count.get(reversed) ?? 0;
-
-        if (reversedCount > 0) {
-            count.set(reversed, reversedCount - 1);
-            res += 2 * word.length;
-        } else {
-            count.set(word, (count.get(word) ?? 0) + 1);
-        }
+  const count: Map<string, number> = new Map();
+  let res = 0;
+
+  for (const word of words) {
+    const reversed = word.split("").reverse().join("");
+    const reversedCount = count.get(reversed) ?? 0;
+
+    if (reversedCount > 0) {
+      count.set(reversed, reversedCount - 1);
+      res += 2 * word.length;
+    } else {
+      count.set(word, (count.get(word) ?? 0) + 1);
     }
+  }
 
-    for (const [word, freq] of count.entries()) {
-        if (word === word.split('').reverse().join('') && freq > 0) {
-            res += word.length;
-            break; // 只能选一个居中的回文串
-        }
+  for (const [word, freq] of count.entries()) {
+    if (word === word.split("").reverse().join("") && freq > 0) {
+      res += word.length;
+      break; // 只能选一个居中的回文串
     }
+  }
 
-    return res;
+  return res;
 }
-```
+```

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@@ -1,8 +1,8 @@
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@@ -2136,6 +2136,23 @@
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@@ -2293,13 +2310,13 @@
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