6kyu Persistent Bugger

题目：

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.

编写一个函数，持久性，它接受一个正参数num并返回它的乘法持久性，这就是你必须将数字中的数字乘上数字直到你到达一个数字的次数。

For example:

persistence(39) === 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit

persistence(999) === 4 // because 9*9*9 = 729, 7*2*9 = 126,
// 1*2*6 = 12, and finally 1*2 = 2

persistence(4) === 0 // because 4 is already a one-digit number

 Sample Tests:

describe('Initial Tests', function () {
  Test.assertEquals(persistence(39),3);
  Test.assertEquals(persistence(4),0);
  Test.assertEquals(persistence(25),2);
  Test.assertEquals(persistence(999),4);
});

 

答案：　

　　　　// 1

              function persistence(num) {

                     var times = 0;

                     num = num.toString();

                     while (num.length > 1) {

                            times++;

                            num = num.split('').map(Number).reduce((a, b) => a * b).toString();

                     }

                     return times;

              }

 

              // 2

              const persistence = num => {

                     return `${num}`.length > 1 ? 1 + persistence(`${num}`.split('').reduce((a, b) => a * +b)) : 0;

                     // +b 隐式类型转换 字符串转数字 +前为空自动默认为求和运算而不是字符串拼接

                     // * - / % 都可以隐式类型转换 字符串转数字

                     // 此处并不需要+  a和b都是字符串 a*b 隐式转换为数字

              }

 

              // 3

              function persistence(num) {

                     var i = 0;

                     for (i; num.toString().length > 1; i++) {

                            num = num.toString().split('').reduce()(function(x,y){return x * y});

                     }

                     return i;

              }

 

              // 4

              const prod = (n) => (n + '').split('').reduce((p,v) => p * v, 1)

              function persistence(num) {

                     let p = 0;

                     while (num > 9) {

                            num = prod (num);

                            p++;

                     }

                     return p;

              }