feat: add solutions to lc problem: No.2131

No.2131.Longest Palindrome by Concatenating Two Letter Words

Author: yanglbme

solution/0000-0099/0001.Two Sum/Solution.c

@@ -1,44 +1,41 @@
-#include <stdlib.h>
 
-int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
-    int capacity = 1;
-    while (capacity < numsSize * 2) capacity <<= 1;
-    int* keys = malloc(capacity * sizeof(int));
-    int* vals = malloc(capacity * sizeof(int));
-    char* used = calloc(capacity, sizeof(char));
-    if (!keys || !vals || !used) {
-        free(keys);
-        free(vals);
-        free(used);
-        *returnSize = 0;
-        return NULL;
-    }
-    for (int i = 0; i < numsSize; ++i) {
-        int x = nums[i];
-        int y = target - x;
-        unsigned int h = (unsigned int) y & (capacity - 1);
-        while (used[h]) {
-            if (keys[h] == y) {
-                int* res = malloc(2 * sizeof(int));
-                res[0] = vals[h];
-                res[1] = i;
-                *returnSize = 2;
-                free(keys);
-                free(vals);
-                free(used);
-                return res;
-            }
-            h = (h + 1) & (capacity - 1);
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     struct ListNode *next;
+ * };
+ */
+
+struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
+    struct ListNode* dummy = (struct ListNode*) malloc(sizeof(struct ListNode));
+    dummy->val = 0;
+    dummy->next = NULL;
+    struct ListNode* curr = dummy;
+    int carry = 0;
+
+    while (l1 != NULL || l2 != NULL || carry != 0) {
+        int sum = carry;
+        if (l1 != NULL) {
+            sum += l1->val;
+            l1 = l1->next;
+        }
+        if (l2 != NULL) {
+            sum += l2->val;
+            l2 = l2->next;
         }
-        unsigned int h2 = (unsigned int) x & (capacity - 1);
-        while (used[h2]) h2 = (h2 + 1) & (capacity - 1);
-        used[h2] = 1;
-        keys[h2] = x;
-        vals[h2] = i;
+
+        carry = sum / 10;
+        int val = sum % 10;
+
+        struct ListNode* newNode = (struct ListNode*) malloc(sizeof(struct ListNode));
+        newNode->val = val;
+        newNode->next = NULL;
+        curr->next = newNode;
+        curr = curr->next;
     }
-    *returnSize = 0;
-    free(keys);
-    free(vals);
-    free(used);
-    return NULL;
-}
+
+    struct ListNode* result = dummy->next;
+    free(dummy);
+    return result;
+}

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

@@ -73,17 +73,17 @@ tags:
 
 ### 方法一：贪心 + 哈希表
 
-我们先用哈希表 `cnt` 统计每个单词出现的次数。
+我们先用一个哈希表 $\textit{cnt}$ 统计每个单词出现的次数。
 
-遍历 `cnt` 中的每个单词 $k$ 以及其出现次数 $v$：
+遍历 $\textit{cnt}$ 中的每个单词 $k$ 以及其出现次数 $v$：
 
-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。
+-   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。
 
-如果 $k$ 中两个字母不同，那么我们要找到一个单词 $k'$，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。
+-   如果 $k$ 中两个字母不同，那么我们要找到一个单词 $k'$，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。
 
 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。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为 `words` 的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为单词的数量。
 
 <!-- tabs:start -->
 
@@ -111,7 +111,7 @@ class Solution {
     public int longestPalindrome(String[] words) {
         Map<String, Integer> cnt = new HashMap<>();
         for (var w : words) {
-            cnt.put(w, cnt.getOrDefault(w, 0) + 1);
+            cnt.merge(w, 1, Integer::sum);
         }
         int ans = 0, x = 0;
         for (var e : cnt.entrySet()) {
@@ -183,6 +183,26 @@ func longestPalindrome(words []string) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function longestPalindrome(words: string[]): number {
+    const cnt = new Map<string, number>();
+    for (const w of words) cnt.set(w, (cnt.get(w) || 0) + 1);
+    let [ans, x] = [0, 0];
+    for (const [k, v] of cnt.entries()) {
+        if (k[0] === k[1]) {
+            x += v & 1;
+            ans += Math.floor(v / 2) * 2 * 2;
+        } else {
+            ans += Math.min(v, cnt.get(k[1] + k[0]) || 0) * 2;
+        }
+    }
+    ans += x ? 2 : 0;
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

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

@@ -73,7 +73,19 @@ Note that "ll" is another longest palindrome that can be created, and
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy + Hash Table
+
+First, we use a hash table $\textit{cnt}$ to count the occurrences of each word.
+
+Iterate through each word $k$ and its count $v$ in $\textit{cnt}$:
+
+-   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.
+
+-   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 $k$, 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.
+
+After the iteration, if $x$ is not empty, we can also place one word in the middle of the palindrome.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of words.
 
 <!-- tabs:start -->
 
@@ -101,7 +113,7 @@ class Solution {
     public int longestPalindrome(String[] words) {
         Map<String, Integer> cnt = new HashMap<>();
         for (var w : words) {
-            cnt.put(w, cnt.getOrDefault(w, 0) + 1);
+            cnt.merge(w, 1, Integer::sum);
         }
         int ans = 0, x = 0;
         for (var e : cnt.entrySet()) {
@@ -173,6 +185,26 @@ func longestPalindrome(words []string) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function longestPalindrome(words: string[]): number {
+    const cnt = new Map<string, number>();
+    for (const w of words) cnt.set(w, (cnt.get(w) || 0) + 1);
+    let [ans, x] = [0, 0];
+    for (const [k, v] of cnt.entries()) {
+        if (k[0] === k[1]) {
+            x += v & 1;
+            ans += Math.floor(v / 2) * 2 * 2;
+        } else {
+            ans += Math.min(v, cnt.get(k[1] + k[0]) || 0) * 2;
+        }
+    }
+    ans += x ? 2 : 0;
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

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

@@ -2,7 +2,7 @@ class Solution {
     public int longestPalindrome(String[] words) {
         Map<String, Integer> cnt = new HashMap<>();
         for (var w : words) {
-            cnt.put(w, cnt.getOrDefault(w, 0) + 1);
+            cnt.merge(w, 1, Integer::sum);
         }
         int ans = 0, x = 0;
         for (var e : cnt.entrySet()) {

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

@@ -0,0 +1,15 @@
+function longestPalindrome(words: string[]): number {
+    const cnt = new Map<string, number>();
+    for (const w of words) cnt.set(w, (cnt.get(w) || 0) + 1);
+    let [ans, x] = [0, 0];
+    for (const [k, v] of cnt.entries()) {
+        if (k[0] === k[1]) {
+            x += v & 1;
+            ans += Math.floor(v / 2) * 2 * 2;
+        } else {
+            ans += Math.min(v, cnt.get(k[1] + k[0]) || 0) * 2;
+        }
+    }
+    ans += x ? 2 : 0;
+    return ans;
+}