javascript 对数字进行因式分解

Factorialize a Number

提问by VADO

I'm taking the freecodecamp course one of the exercises it's to create a Factorializefunction, I know there is several ways to do it just not sure what this one keeps returning 5

我正在参加 freecodecamp 课程，其中一项练习是创建一个Factorialize函数，我知道有几种方法可以做到，只是不确定这个不断返回什么5

function factorialize(num) {
    var myMax = num;
    var myCounter = 1;
    var myTotal = 0;

    for (i = 0; i>= myMax; i++) {
        num = myCounter * (myCounter + 1);
        myCounter++;
    }
    return num;
}

factorialize(5);

回答by Andro

This is a recursive solution of your problem:

这是您问题的递归解决方案：

function factorialize(num) {
    if(num <= 1) {
        return num
    } else {
        return num * factorialize(num-1) 
    }
}

factorialize(5)

This is the iterative solution:

这是迭代解决方案：

function factorialize(num) {
    var cnt = 1;
    for (var i = 1; i <= num ; i++) {
        cnt *= i;
    }
    return cnt;
}

factorialize(5)

with argument 5, it will return the 5! or 120.

使用参数 5，它将返回 5！或 120。

回答by Koleen BP

My solution in compliance with convention for empty product

我的符合空产品公约的解决方案

function factorializer(int) {
    if (int <= 1) {
        return 1;
    } else {
        return int * factorializer(int - 1);
    }
}

回答by Hans Mikesen

To answer your question, why your function is returning 5: Your function never reaches the inner part of the for-loop because your testing if i is greater than myMax instead of less than. So you are just returning your input parameter which is five.

要回答您的问题，为什么您的函数返回 5：您的函数永远不会到达 for 循环的内部部分，因为您测试 i 是否大于 myMax 而不是小于。所以你只是返回你的输入参数，它是 5。

But the loop does not calculate the factorial of num, it only multiplies (num+1) with (num+2);

但是循环不计算 num 的阶乘，它只是将 (num+1) 与 (num+2) 相乘；

回答by Ahmad Saleh

Here is another way to solve this challenge and I know it is neither the shortest nor the easiest but it is still a valid way.

这是解决这一挑战的另一种方法，我知道它既不是最短的也不是最简单的，但它仍然是一种有效的方法。

function factorialiaze(num){
var myArr = []; //declaring an array.
if(num === 0 || num === 1){
return 1;
}
if (num < 0){ //for negative numbers.
return "N/A";
}
for (var i = 1; i <= num; i++){ // creating an array.
     myArr.push(i);
}
// Reducing myArr to a single value via .reduce:
num = myArr.reduce(function(a,b){
     return a * b;
 });
return num;
}
factorialiaze(5);

回答by Imran Khan

Try this function

试试这个功能

const factorialize = (num) =>  num === 0 ? 1 : num * factorialize(num-1)

Use it like this:

像这样使用它：

factorialize(5) // returns 120

回答by Marina ES

Try this :

试试这个 ：

function factorialize(num) {
  var value = 1;
    if(num === 1 || num ===0) {
      return value;
    } else {
      for(var i = 1; i<num; i++) {
        value *= i;
      }
      return num * value;
  }
}
factorialize(5);

回答by Nina Scholz

Maybe you consider another approach.

也许你考虑另一种方法。

This solution features a very short - cut to show what is possible to get with an recursive style and a implicit type conversion:

此解决方案具有一个非常快捷的功能，以展示使用递归样式和隐式类型转换可能获得的结果：

function f(n) { return +!~-n || n * f(n - 1); }

• + convert to number
• ! not
• ~ not bitwise
• - negative

• + 转换为数字
• ！不是
• ~ 不是按位
• - 消极的

function f(n) { return +!~-n || n * f(n - 1); }

var i;
for (i = 1; i < 20; i++) {
    console.log(f(i));
}

.as-console-wrapper { max-height: 100% !important; top: 0; }