Question

我的代码有效，但计数不正确。例如，它正在考虑 22 和 208 作为happy numbers，但它们不是。{{3}}。乘法结果也是错误的：在 42 之后，它不是 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')
      }
    }
}

Answer 2

您可以使用较少的变量稍微更改代码，并使用break退出。

基本上你不需要迭代获得正方形，然后另一个循环来获得正方形的总和。使用数组a及其以前的值是错误结果的原因。

＆＃13;

function isHappy(x) {
    var b = x,
        str,
        dejavue = [],
        sum,
        result;

    while (true) {
        str = b.toString();
        sum = 0;
        for (var i = 0; i < str.length; i++) {
            sum += str[i] * str[i];
        }
        if (dejavue.indexOf(sum) !== -1) {
            result = " is not happy";
            break;
        }
        dejavue.push(sum);
        if (sum === 1) {
            result = " is happy";
            break;
        }
        b = sum;
    }
    return dejavue.join('>') + ": " + x + result;
}

console.log(isHappy(22));
console.log(isHappy(7));

＆＃13;

Javascript Happy Number检测错误

2 个答案: