Javascript Happy Number 检测错误

Question

我的代码可以工作，但计数不正确。例如，它将 22 和 208 视为 happy numbers ，但他们不是。乘法结果也是错误的:42之后不是20，而是45

    function isHappy(x){
   var b = x,
       str = b.toString(), 
       strlen = str.length, 
       a = [], 
       dejavue = [], 
       sum = 0, 
       isOne = false, 
       result, 
       whilenumber = -1;

   while (isOne == false){
   whilenumber++;
   for (var i=0;i<strlen;i++){
       var ms = parseInt(str[i]);
       a[i] = ms*ms;
       //a[i] = Math.pow(ms, 2);   
   }
   if (a.length>1){
       for (var i=0;i<a.length;i++){
       sum = sum + a[i];  
       }   
   } else { sum = a[0];}
   var h = dejavue.indexOf(sum);
   if (h==-1) {dejavue[whilenumber] = sum;} else {
       result = " is not happy";
       break;   
   }
   if (sum!=1){
   b = sum;
   sum = 0;
   str = b.toString();
   strlen = str.length;
   }else{
   result = " is happy";
   isOne = true;
   }
}
   return dejavue+"_"+x+result;   
}

Answer 1

var l = 50;
a=n=>{if((r=[...(n+'')].map(t=>t*t).reduce((a,b)=>a+b))==1){alert('happy')}else{if(l){l--;a(r)}else{alert('unhappy')}}
}

a(22)

/* explanation */

// how many times we want our function to run ultil decide is a unhappy number. This is to avoid infinite loop
var l = 50;

// parameter n = positive integer
n=>{
  if(
    //convert integer to string and split each char into an array
    (r=[...(n+'')]
    // get each separated number its square
    .map(t=>t*t)
    // sum squares
    .reduce((a,b)=>a+b)) 
    //if result is equal to one
    == 1
  ){
      // we alert happy
      alert('happy')
    }else{
      //otherwise we check if our loop counter (l) aint done looping
      if(l){
        // if it does, then we substract 1 from the counter and call this same function with the result of the above operation
        l--;a(r)
      }else{
      // otherwise is a unhappy number
        alert('unhappy')
      }
    }
}