codewars--js--Happy numbers++无穷大判断

问题描述：

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, and 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 numbers, while those that do not end in 1 are unhappy numbers (or sad numbers) (Wikipedia).

For example number 7 is happy because after a number of steps the computed sequence ends up with a 1: 7, 49, 97, 130, 10, 1

while 3 is not, and would give us an infinite sequence: 3, 9, 81, 65, 61, 37, 58, 89, 145, 42, 20, 4, 16,3 7, 58, 89, 145, 42, 20, 4, 16, 37, ...

Write a function that takes n as parameter and return true if and only if n is an happy number.

Happy coding!

我的思路：

当求得sum=1时结束循环（sum=当前数各个位的平方之和）。对于有的数不是happy数，这个while就得一直做下去，设置一个flag判断是都为无穷大数（Number.POSITIVE_INFINITY），若是，则直接返回false。但是这样仍然超时。

我的答案：

function isHappy(n) {
  // Good Luck
  var sum=0;
  var flag=1;
  while(sum!=1){
    var a=n.toString().split("");
    for(var i=0;i<a.length;i++){
      sum=sum+Math.pow((+a[i]),2);
    }
    n=sum;
    flag+=1;
    if(flag=="Infinity"){return false;}
  }
  return true;
}

优秀答案：

（1）通过判断n是不是循环出现，跳出while循环

function isHappy(n) {
  let arr = []
  while (n !== 1 && arr.indexOf(n) === -1) {
    arr.push(n);
    n = n.toString().split('').map(x => Math.pow(Number(x), 2)).reduce((p, n) => p + n, 0);
  }
  return n ==1?true:false;
}

哈哈哈！