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Commit 7381ede

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feat: add solutions to lc problem: No.2131 (doocs#4428)
No.2131.Longest Palindrome by Concatenating Two Letter Words
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‎solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/README.md‎

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### 方法一:贪心 + 哈希表
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我们先用哈希表 `cnt` 统计每个单词出现的次数。
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我们先用一个哈希表 $\textit{cnt}$ 统计每个单词出现的次数。
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遍历 `cnt` 中的每个单词 $k$ 以及其出现次数 $v$:
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遍历 $\textit{cnt}$ 中的每个单词 $k$ 以及其出现次数 $v$:
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如果 $k$ 中两个字母相同,那么我们可以将 $\left \lfloor \frac{v}{2} \right \rfloor \times 2$ 个 $k$ 连接到回文串的前后,此时如果 $k$ 还剩余一个,那么我们可以先记录到 $x$ 中。
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-如果 $k$ 中两个字母相同,那么我们可以将 $\left \lfloor \frac{v}{2} \right \rfloor \times 2$ 个 $k$ 连接到回文串的前后,此时如果 $k$ 还剩余一个,那么我们可以先记录到 $x$ 中。
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如果 $k$ 中两个字母不同,那么我们要找到一个单词 $k',ドル使得 $k'$ 中的两个字母与 $k$ 相反,即 $k' = k[1] + k[0]$。如果 $k'$ 存在,那么我们可以将 $\min(v, cnt[k'])$ 个 $k$ 连接到回文串的前后。
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-如果 $k$ 中两个字母不同,那么我们要找到一个单词 $k',ドル使得 $k'$ 中的两个字母与 $k$ 相反,即 $k' = k[1] + k[0]$。如果 $k'$ 存在,那么我们可以将 $\min(v, \textit{cnt}[k'])$ 个 $k$ 连接到回文串的前后。
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遍历结束后,如果 $x$ 不为空,那么我们还可以将一个单词连接到回文串的中间。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为 `words` 的长度。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为单词的数量。
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<!-- tabs:start -->
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@@ -111,7 +111,7 @@ class Solution {
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public int longestPalindrome(String[] words) {
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Map<String, Integer> cnt = new HashMap<>();
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for (var w : words) {
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cnt.put(w, cnt.getOrDefault(w, 0) +1);
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cnt.merge(w, 1, Integer::sum);
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}
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int ans = 0, x = 0;
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for (var e : cnt.entrySet()) {
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}
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```
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#### TypeScript
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```ts
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function longestPalindrome(words: string[]): number {
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const cnt = new Map<string, number>();
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for (const w of words) cnt.set(w, (cnt.get(w) || 0) + 1);
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let [ans, x] = [0, 0];
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for (const [k, v] of cnt.entries()) {
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if (k[0] === k[1]) {
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x += v & 1;
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ans += Math.floor(v / 2) * 2 * 2;
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} else {
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ans += Math.min(v, cnt.get(k[1] + k[0]) || 0) * 2;
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}
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}
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ans += x ? 2 : 0;
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Greedy + Hash Table
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First, we use a hash table $\textit{cnt}$ to count the occurrences of each word.
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Iterate through each word $k$ and its count $v$ in $\textit{cnt}$:
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- If the two letters in $k$ are the same, we can concatenate $\left \lfloor \frac{v}{2} \right \rfloor \times 2$ copies of $k$ to the front and back of the palindrome. If there is one $k$ left, we can record it in $x$ for now.
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- If the two letters in $k$ are different, we need to find a word $k'$ such that the two letters in $k'$ are the reverse of $k,ドル i.e., $k' = k[1] + k[0]$. If $k'$ exists, we can concatenate $\min(v, \textit{cnt}[k'])$ copies of $k$ to the front and back of the palindrome.
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After the iteration, if $x$ is not empty, we can also place one word in the middle of the palindrome.
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The time complexity is $O(n),ドル and the space complexity is $O(n),ドル where $n$ is the number of words.
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<!-- tabs:start -->
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public int longestPalindrome(String[] words) {
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Map<String, Integer> cnt = new HashMap<>();
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for (var w : words) {
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cnt.put(w, cnt.getOrDefault(w, 0) +1);
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cnt.merge(w, 1, Integer::sum);
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}
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int ans = 0, x = 0;
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for (var e : cnt.entrySet()) {
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}
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```
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#### TypeScript
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```ts
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function longestPalindrome(words: string[]): number {
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const cnt = new Map<string, number>();
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for (const w of words) cnt.set(w, (cnt.get(w) || 0) + 1);
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let [ans, x] = [0, 0];
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for (const [k, v] of cnt.entries()) {
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if (k[0] === k[1]) {
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x += v & 1;
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ans += Math.floor(v / 2) * 2 * 2;
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} else {
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ans += Math.min(v, cnt.get(k[1] + k[0]) || 0) * 2;
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}
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}
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ans += x ? 2 : 0;
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/Solution.java‎

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public int longestPalindrome(String[] words) {
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Map<String, Integer> cnt = new HashMap<>();
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for (var w : words) {
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cnt.put(w, cnt.getOrDefault(w, 0) + 1);
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cnt.merge(w, 1, Integer::sum);
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}
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int ans = 0, x = 0;
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for (var e : cnt.entrySet()) {
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function longestPalindrome(words: string[]): number {
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const cnt = new Map<string, number>();
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for (const w of words) cnt.set(w, (cnt.get(w) || 0) + 1);
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let [ans, x] = [0, 0];
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for (const [k, v] of cnt.entries()) {
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if (k[0] === k[1]) {
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x += v & 1;
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ans += Math.floor(v / 2) * 2 * 2;
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} else {
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ans += Math.min(v, cnt.get(k[1] + k[0]) || 0) * 2;
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}
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}
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ans += x ? 2 : 0;
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return ans;
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}

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