chore: add daily leetcode post 2131. 连接两字母单词得到的最长回文串

Author: longsizhuo

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

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+---
+title: 2131. 连接两字母单词得到的最长回文串.md
+date: '2025/5/25-2:33'
+tags:
+  - - Python
+  - - Answer
+abbrlink: 9fa195e5
+---
+
+# QUESTION：
+
+[2131. 连接两字母单词得到的最长回文串.md](https://leetcode.cn/problems/longest-palindrome-by-concatenating-two-letter-words/description/?envType=daily-question&envId=2025-05-25)
+
+# My Think：
+
+The idea comes from the brilliant Ling-nc.
+
+We build a hash map to count the occurrences of each word.
+
+For each word, we check whether its reverse exists in the hash map.
+If it does, they can form a palindrome pair—we update the result and decrease the corresponding count.
+If the reverse does not exist, we increment the count for the current word.
+Finally, we check if there’s any palindromic word that can be used as the center of the palindrome.
+
+Example:
+
+1. Input: ["lc", "cl", "gg", "gg"]
+
+2. "lc" has no pair → stored in the map: { "lc": 1 }
+
+3. "cl" finds "lc" exists → use "lc" + "cl" as a pair → res += 4 → map updated to { "lc": 0 }
+
+4. "gg" has no pair → stored in the map: { "lc": 0, "gg": 1 }
+
+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.
+
+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.
+
+
+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 }
+
+3. "cl" 发现 "lc" 存在 → 成对使用 "lc"+"cl" → res += 4 → map 更新为 { "lc": 0 }
+
+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。 
+
+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。
+# Code：
+
+```python
+from collections import defaultdict
+from typing import List
+class Solution:
+    def longestPalindrome(self, words: List[str]) -> int:
+        count = defaultdict(int)
+        res = 0
+        for word in words:
+            if count[word[::-1]] > 0:
+                count[word[::-1]] -= 1
+                res += 4
+            else:
+                count[word] += 1
+        for key, value in count.items():
+            if key[0] == key[1] and value > 0:
+                res += 2
+                break
+        return res
+```
+
+```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);
+        }
+    }
+
+    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;
+}
+```