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docs: 更新多个题解文件以匹配 leetcoder.net 的格式
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‎en/1-1000/202-happy-number.md

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# 202. Happy Number - Best Practices of LeetCode Solutions
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Original link: [leetcoder.net - Fucking Good LeetCode Solutions](https://leetcoder.net/en/leetcode/202-happy-number)
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# 202. Happy Number - Fucking Good LeetCode Solutions
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LeetCode link: [202. Happy Number](https://leetcode.com/problems/happy-number), difficulty: **Easy**.
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## LeetCode description of "202. Happy Number"

‎en/1-1000/28-find-the-index-of-the-first-occurrence-in-a-string.md

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# 28. Find the Index of the First Occurrence in a String - Best Practices of LeetCode Solutions
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Original link: [leetcoder.net - Fucking Good LeetCode Solutions](https://leetcoder.net/en/leetcode/28-find-the-index-of-the-first-occurrence-in-a-string)
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# 28. Find the Index of the First Occurrence in a String - Fucking Good LeetCode Solutions
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LeetCode link: [28. Find the Index of the First Occurrence in a String](https://leetcode.com/problems/find-the-index-of-the-first-occurrence-in-a-string), difficulty: **Easy**.
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## LeetCode description of "28. Find the Index of the First Occurrence in a String"
@@ -100,7 +102,27 @@ var strStr = function (haystack, needle) {
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# Welcome to create a PR to complete the code of this language, thanks!
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```
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## C, Kotlin, Swift, Rust or other languages
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## C
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```c
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Kotlin
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```kotlin
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Swift
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```swift
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Rust
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```rust
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Other languages
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```

‎en/1-1000/349-intersection-of-two-arrays.md

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# 349. Intersection of Two Arrays - Best Practices of LeetCode Solutions
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Original link: [leetcoder.net - Fucking Good LeetCode Solutions](https://leetcoder.net/en/leetcode/349-intersection-of-two-arrays)
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# 349. Intersection of Two Arrays - Fucking Good LeetCode Solutions
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LeetCode link: [349. Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays), difficulty: **Easy**.
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## LeetCode description of "349. Intersection of Two Arrays"
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2. When traversing the other array, if an element is found to already exist in the `set`, it means that the element belongs to the intersection, and the element should be added to the `results`.
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3. The `results` is also of `set` type because duplicate removal is required.
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## Steps
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1. Convert one of the arrays to a `set`. The elements are unique in a `set`.
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```javascript
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let num1Set = new Set(nums1)
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```
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2. When traversing the other array, if an element is found to already exist in the `set`, it means that the element belongs to the intersection, and the element should be added to the `results`.
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```javascript
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let results = new Set()
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for (const num of nums2) {
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if (num1Set.has(num)) {
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results.add(num)
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}
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}
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return Array.from(results)
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```
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## Complexity
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* Time: `O(N)`.
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* Space: `O(N)`.
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* Time: `O(n)`.
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* Space: `O(n)`.
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## Java
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```java

‎en/1-1000/383-ransom-note.md

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# 383. Ransom Note - Best Practices of LeetCode Solutions
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# 383. Ransom Note - Fucking Good LeetCode Solutions
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LeetCode link: [383. Ransom Note](https://leetcode.com/problems/ransom-note),
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[383. 赎金信](https://leetcode.cn/problems/ransom-note)
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return true
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```
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## 复杂度
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* 时间：`O(n)`。
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* 空间：`O(n)`。
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## Java
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```java
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class Solution {
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public boolean canConstruct(String ransomNote, String magazine) {
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var charToCount = new HashMap<Character, Integer>();
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for (var character : magazine.toCharArray()) {
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charToCount.put(character, charToCount.getOrDefault(character, 0) + 1);
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}
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for (var character : ransomNote.toCharArray()) {
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charToCount.put(character, charToCount.getOrDefault(character, 0) - 1);
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if (charToCount.get(character) < 0) {
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return false;
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}
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}
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return true;
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}
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}
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```
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## Python
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```python
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# from collections import defaultdict
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class Solution:
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def canConstruct(self, ransomNote: str, magazine: str) -> bool:
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char_to_count = defaultdict(int)
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for char in magazine:
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char_to_count[char] += 1
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for char in ransomNote:
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char_to_count[char] -= 1
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if char_to_count[char] < 0:
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return False
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return True
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```
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## JavaScript
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```javascript
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var canConstruct = function (ransomNote, magazine) {
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const charToCount = new Map()
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for (const character of magazine) {
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charToCount.set(character, (charToCount.get(character) || 0) + 1)
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}
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for (const character of ransomNote) {
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charToCount.set(character, (charToCount.get(character) || 0) - 1)
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if (charToCount.get(character) < 0) {
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return false
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}
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}
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return true
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};
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```
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## C#
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```c#
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public class Solution
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{
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public bool CanConstruct(string ransomNote, string magazine)
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{
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var charToCount = new Dictionary<char, int>();
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foreach (char character in magazine)
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charToCount[character] = charToCount.GetValueOrDefault(character, 0) + 1;
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foreach (char character in ransomNote)
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{
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charToCount[character] = charToCount.GetValueOrDefault(character, 0) - 1;
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if (charToCount[character] < 0)
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{
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return false;
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}
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}
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return true;
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}
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}
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```
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## Other languages
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```java
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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原文链接：[leetcoder.net - Fucking Good LeetCode Solutions](https://leetcoder.net/en/leetcode/383-ransom-note)

‎en/1-1000/454-4sum-ii.md

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# 454. 4Sum II - Best Practices of LeetCode Solutions
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LeetCode link: [454. 4Sum II](https://leetcode.com/problems/problem), difficulty: **Medium**.
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原文链接：[leetcoder.net - Fucking Good LeetCode Solutions](https://leetcoder.net/en/leetcode/454-4sum-ii)
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# 454. 4Sum II - Fucking Good LeetCode Solutions
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LeetCode link: [454. 4Sum II](https://leetcode.com/problems/4sum-ii), difficulty: **Medium**.
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## LeetCode description of "454. 4Sum II"
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Given four integer arrays `nums1`, `nums2`, `nums3`, and `nums4` all of length `n`, return the number of tuples `(i, j, k, l)` such that:

‎zh/1-1000/202-happy-number.md

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原文链接：[leetcoder.net - 力扣题解最佳实践](https://leetcoder.net/zh/leetcode/202-happy-number)
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# 202. 快乐数 - 力扣题解最佳实践
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力扣链接：[202. 快乐数](https://leetcode.cn/problems/happy-number) ，难度：**简单**。
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## 力扣“202. 快乐数”问题描述
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## 力扣"202. 快乐数"问题描述
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编写一个算法来判断一个数 `n` 是不是快乐数。
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「**快乐数**」 定义为：
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「快乐数」定义为：
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- 对于一个正整数，每一次将该数替换为它每个位置上的数字的平方和。
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- 然后重复这个过程直到这个数变为 `1`，也可能是 **无限循环** 但始终变不到 `1`。
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- 如果这个过程 **结果为** `1`，那么这个数就是快乐数。
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- 如果 `n` 是 *快乐数* 就返回 `true` ；不是，则返回 `false` 。
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- 然后重复这个过程直到这个数变为 1，也可能是 **无限循环** 但始终变不到 1。
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- 如果 **可以变为** 1，那么这个数就是快乐数。
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如果 `n` 是快乐数就返回 `true` ；不是，则返回 `false` 。
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### [示例 1]
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**输入**: `n = 19`
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**解释**:
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```
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1**2 + 9**2 = 82
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8**2 + 2**2 = 68
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6**2 + 8**2 = 100
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1**2 + 0**2 + 0**2 = 1
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1^2 + 9^2 = 82
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8^2 + 2^2 = 68
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6^2 + 8^2 = 100
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1^2 + 0^2 + 0^2 = 1
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```
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### [示例 2]
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**输出**: `false`
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### [约束]
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- `1 <= n <= 2**31 - 1`
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- `1 <= n <= 2^31 - 1`
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## 思路
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1. 递归调用`isHappy(n)`比较方便，每次只需要生成新的`n`作为参数。
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2. 如果`n`已经出现过了，说明进入了循环，`return false`。可以用`Set`保存已经出现过的`n`。
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1. 递归调用 `isHappy(n)` 更方便。每次只需要生成新的 `n` 作为参数。
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2. 如果 `n` 已经出现过，说明进入了循环，`return false`。可以用 `Set` 保存出现过的 `n`。
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## 步骤
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1. 生成新的`n`作为`isHappy()`的参数。
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1. 生成新的 `n` 作为 `isHappy(n)` 的参数。
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```javascript
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let sum = 0
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return isHappy(sum)
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```
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2. 如果`n`已经出现过了，说明进入了循环，`return false`。可以用`Set`保存已经出现过的`n`。
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2. 如果 `n` 已经出现过，说明进入了循环，`return false`。可以用 `Set` 保存出现过的 `n`。
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```javascript
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var isHappy = function (n, appearedNums) {
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var isHappy = function (n, appearedNums) { // 0
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```
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## 复杂度
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* 时间：`O(log N)`。
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* 空间：`O(log N)`。
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* 时间: `O(log N)`.
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* 空间: `O(log N)`.
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## Java
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```java

‎zh/1-1000/28-find-the-index-of-the-first-occurrence-in-a-string.md

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原文链接：[leetcoder.net - 力扣题解最佳实践](https://leetcoder.net/zh/leetcode/28-find-the-index-of-the-first-occurrence-in-a-string)
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# 28. 找出字符串中第一个匹配项的下标 - 力扣题解最佳实践
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力扣链接：[28. 找出字符串中第一个匹配项的下标](https://leetcode.cn/problems/find-the-index-of-the-first-occurrence-in-a-string) ，难度：**简单**。
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## 力扣“28. 找出字符串中第一个匹配项的下标”问题描述
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## 力扣"28. 找出字符串中第一个匹配项的下标"问题描述
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给你两个字符串 `haystack` 和 `needle` ，请你在 `haystack` 字符串中找出 `needle` 字符串的第一个匹配项的下标（下标从 `0` 开始）。如果 `needle` 不是 `haystack` 的一部分，则返回 `-1` 。
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### [示例 1]

‎zh/1-1000/383-ransom-note.md

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# 383. 赎金信 - 力扣题解最佳实践
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力扣链接：[383. 赎金信](https://leetcode.cn/problems/ransom-note), 难度：**简单**。
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力扣链接：[383. 赎金信](https://leetcode.cn/problems/ransom-note) ，难度：**简单**。
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## 力扣“383. 赎金信”问题描述
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## 力扣"383. 赎金信"问题描述
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给你两个字符串：`ransomNote` 和 `magazine` ，判断 `ransomNote` 能不能由 `magazine` 里面的字符构成。
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**输出**: `true`
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### [约束]
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- `1 <= ransomNote.length, magazine.length <= 10^5`
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- `1 <= ransomNote.length, magazine.length <= 100000`
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- `ransomNote` 和 `magazine` 由小写英文字母组成
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## 思路
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1. 本题等同于求`magazine`是否能包含`ransomNote`中的所有字符。
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2. 先对`magazine`进行统计，得出每个字符对应的字数，结果存储在`Map`中。每一次都是一个加一的操作。
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3. 下一步做什么？
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<details><summary>点击查看答案</summary><p>遍历`ransomNote`，对当前字符对应的数量进行减一操作（反向操作）。如果某个字符的数量小于0，则返回`false`。</p></details>
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2. 先对`magazine`进行字符和字数统计，结果存储在`Map`中。
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3. 然后，遍历`ransomNote`，并对`Map`中的数据进行反向操作。如果某个字符的字数小于0，则返回`false`。
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## 步骤
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## 复杂度
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- 时间复杂度: `O(N)`.
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- 空间复杂度: `O(N)`.
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- 时间复杂度: `O(n)`.
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- 空间复杂度: `O(n)`.
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## Java
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}
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```
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## C++
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```cpp
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Go
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```go
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Ruby
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```ruby
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# Welcome to create a PR to complete the code of this language, thanks!
188+
```
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## C
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```c
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// Welcome to create a PR to complete the code of this language, thanks!
194+
```
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## Kotlin
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```kotlin
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Swift
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```swift
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// Welcome to create a PR to complete the code of this language, thanks!
206+
```
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## Rust
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```rust
211+
// Welcome to create a PR to complete the code of this language, thanks!
212+
```
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## Other languages
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```java
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```
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// Welcome to create a PR to complete the code of this language, thanks!
177218
```

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