# LeetCode Notes: Happy Number

## Question

Write an algorithm to determine if a number `n`

is happy.

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.

Return `true`

*if* `n`

*is a happy number, and* `false`

*if not*.

**Example 1:**

Input:n = 19

Output:true

Explanation:1

^{2}+ 9^{2}= 828

^{2}+ 2^{2}= 686

^{2}+ 8^{2}= 1001

^{2}+ 0^{2}+ 0^{2}= 1

**Example 2:**

Input:n = 2

Output:false

**Constraints:**

- 1 <= n <= 2
^{31}- 1

## Solution

**Analysis:**

There are two situations

- Loop the sum of squares to get the result 1, which is true
- Infinite loop, no result 1 is obtained. At this time, use the fast and slow pointers to determine whether there are two pointers meeting to determine whether it is in an infinite loop

**Code:**

```
/**
* @param {number} n
* @return {boolean}
*/
var isHappy = function(n) {
let after = n // current value
let forward = nextNum(n) // step forward
// Exit the loop in the following two cases
// forward == after: fast and slow pointers meet
// forward == 1: got the result 1
while(forward != after && forward != 1){
after = nextNum(after) // Slow pointer goes one step forward
forward = nextNum(nextNum(forward)) // Fast pointer goes two steps forward
}
// return result
return forward == 1
// Get the next number, that is, the sum of the squares of the numbers at each position
function nextNum(n){
return (n +'').split('').map(a => a ** 2).reduce((x,y)=> x + y)
}
};
```

## Comments