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‎en/1-1000/202-happy-number.md

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@@ -6,11 +6,11 @@ Write an algorithm to determine if a number `n` is happy.
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A **happy number** is a number defined by the following process:
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* Starting with any positive integer, replace the number by the sum of the squares of its digits.
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* Repeat the process until the number equals 1 (where it will stay), or it **loops endlessly in a cycle** which does not include 1.
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* Those numbers for which this process **ends in 1** are happy.
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- Starting with any positive integer, replace the number by the sum of the squares of its digits.
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- Repeat the process until the number equals 1 (where it will stay), or it **loops endlessly in a cycle** which does not include 1.
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- Those numbers for which this process **ends in 1** are happy.
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Return `true` if `n` is _a happy number_, and `false` if not.
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Return `true` if `n` is *a happy number*, and `false` if not.
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### [Example 1]
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**Input**: `n = 19`
@@ -19,10 +19,10 @@ Return `true` if `n` is _a happy number_, and `false` if not.
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**Explanation**
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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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### [Example 2]
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**Output**: `false`
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### [Constraints]
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- `1 <= n <= 2**31 - 1`
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- `1 <= n <= 2^31 - 1`
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## Intuition
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1. It is more convenient to call `isHappy(n)` recursively. You only need to generate a new `n` as a parameter each time.
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2. If `n` has already appeared, it means that the loop has been entered, and `return false`. You can use `Set` to save the `n` that has appeared.
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## Steps
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1. Generate a new `n` as the `isHappy(n)` parameter.
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```javascript
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let sum = 0
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for (const digit of n.toString()) {
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sum += Math.pow(Number(digit), 2)
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}
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// omitted code
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1. Generate a new `n` as the `isHappy(n)` parameter.
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return isHappy(sum)
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```
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```javascript
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let sum = 0
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for (const digit of n.toString()) {
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sum += Math.pow(Number(digit), 2)
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}
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// omitted code
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return isHappy(sum)
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```
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2. If `n` has already appeared, it means that the loop has been entered, and `return false`. You can use `Set` to save the `n` that has appeared.
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```javascript
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var isHappy = function (n, appearedNums) { // 0
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appearedNums ||= new Set() // 1
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let sum = 0
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for (const digit of n.toString()) {
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sum += Math.pow(Number(digit), 2)
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}
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if (sum == 1) {
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return true
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}
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if (appearedNums.has(sum)) { // 2
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return false
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}
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appearedNums.add(sum) // 3
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return isHappy(sum, appearedNums)
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};
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```
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```javascript
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var isHappy = function (n, appearedNums) { // 0
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appearedNums ||= new Set() // 1
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let sum = 0
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for (const digit of n.toString()) {
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sum += Math.pow(Number(digit), 2)
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}
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if (sum == 1) {
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return true
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}
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if (appearedNums.has(sum)) { // 2
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return false
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}
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appearedNums.add(sum) // 3
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return isHappy(sum, appearedNums)
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};
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```
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## Complexity
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* Time: `O(log N)`.

‎zh/1-1000/202-happy-number.md

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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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- 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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### [示例 1]
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**输入**: `n = 19`
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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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## 步骤
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1. 生成新的`n`作为`isHappy()`的参数。
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```javascript
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let sum = 0
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for (const digit of n.toString()) {
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sum += Math.pow(Number(digit), 2)
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}
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// omitted code
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1. 生成新的`n`作为`isHappy()`的参数。
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return isHappy(sum)
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```
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```javascript
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let sum = 0
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for (const digit of n.toString()) {
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sum += Math.pow(Number(digit), 2)
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}
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// omitted code
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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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```javascript
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var isHappy = function (n, appearedNums) {
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appearedNums ||= new Set() // 1
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let sum = 0
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for (const digit of n.toString()) {
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sum += Math.pow(Number(digit), 2)
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}
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if (sum == 1) {
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return true
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}
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if (appearedNums.has(sum)) { // 2
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return false
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}
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appearedNums.add(sum) // 3
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return isHappy(sum, appearedNums)
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};
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```
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```javascript
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var isHappy = function (n, appearedNums) {
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appearedNums ||= new Set() // 1
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let sum = 0
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for (const digit of n.toString()) {
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sum += Math.pow(Number(digit), 2)
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}
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if (sum == 1) {
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return true
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}
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if (appearedNums.has(sum)) { // 2
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return false
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}
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appearedNums.add(sum) // 3
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return isHappy(sum, appearedNums)
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};
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```
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## 复杂度
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* 时间：`O(log N)`。

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