A**happy number** is a number defined by the following process:

* Starting with any positive integer, replace the number by the sum of the squares of its digits.

* Repeat the process until the number equals 1 (where it will stay), or it**loops endlessly in a cycle** which does not include 1.

* Those numbers for which this process**ends in 1** are happy.

- Starting with any positive integer, replace the number by the sum of the squares of its digits.

- Repeat the process until the number equals 1 (where it will stay), or it**loops endlessly in a cycle** which does not include 1.

- Those numbers for which this process**ends in 1** are happy.

Return`true` if`n` is_a happynumber_, and`false` if not.

Return`true` if`n` is*a happynumber*, and`false` if not.

**Input**:`n = 19`

**Explanation**

**Output**:`false`

1. It is more convenient to call`isHappy(n)` recursively. You only need to generate a new`n` as a parameter each time.

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.

1. Generate a new`n` as the`isHappy(n)` parameter.

let sum=0

for (constdigitofn.toString()) {

sum+=Math.pow(Number(digit),2)

}

returnisHappy(sum)

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.

varisHappy=function (n,appearedNums) {// 0

appearedNums||=newSet()// 1

let sum=0

for (constdigitofn.toString()) {

sum+=Math.pow(Number(digit),2)

}

if (sum==1) {

}

if (appearedNums.has(sum)) {// 2

}

appearedNums.add(sum)// 3

returnisHappy(sum, appearedNums)

};

* Time:`O(log N)`.

「**快乐数**」 定义为：

* 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。

* 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`。

* 如果这个过程**结果为**`1`，那么这个数就是快乐数。

* 如果`n` 是_快乐数_ 就返回`true` ；不是，则返回`false` 。

- 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。

- 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`。

- 如果这个过程**结果为**`1`，那么这个数就是快乐数。

- 如果`n` 是*快乐数* 就返回`true` ；不是，则返回`false` 。

**输入**:`n = 19`

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`。

1. 生成新的`n`作为`isHappy()`的参数。

let sum=0

for (constdigitofn.toString()) {

sum+=Math.pow(Number(digit),2)

}

returnisHappy(sum)

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`。

varisHappy=function (n,appearedNums) {

appearedNums||=newSet()// 1

let sum=0

for (constdigitofn.toString()) {

sum+=Math.pow(Number(digit),2)

}

if (sum==1) {

}

if (appearedNums.has(sum)) {// 2

}

appearedNums.add(sum)// 3

returnisHappy(sum, appearedNums)

};

* 时间：`O(log N)`。