javascript - 有条件的 : If all numbers in array are divisors without remainder

Question

我的问题是:

Print all numbers from low to high. If any number being printed is divisible by any divisor number in arr, print the number + the word ¨one_match¨ If the number being printed is divisible by ALL the numbers in the array, it should output the number + “all_match”.

在下面的函数中，我似乎能够为数组中的数字之一(通过遍历数组)和基本情况(没有一个是可整除的)生成条件。我正在寻求中间情况的帮助(如果所有情况都可以整除)，似乎内部循环不适合这个...额外的 for...in 并没有真正与数组相提并论...完成功能的任何想法？ (或完全优化它)

function check(arr, low, high) { 
  for (var i = low; i <= high; i++) {
    for (var j = 0; j < arr.length; j++)
      if (i % arr[j] === 0)
        console.log(i + ' one_match')
        //  trouble here ->   else if (i % arr[j] && ???)
      else
        console.log(i);
  }
};

check([2,3],1,7);

Answer 1

您可以只计算匹配项并查看内部循环后的结果:

function check(arr, low, high) { 
  for (var i = low; i <= high; i++) {
    var count = 0;
    for (var j = 0; j < arr.length; j++) {
      if (i % arr[j] === 0) {
        count++;
      }
    }
    if (count === arr.length) {
      console.log(i + ' all_match');
    } else if (count > 0) {
      console.log(i + ' one_match');
    } else {
      console.log(i);
    }
  }
};

check([2,3],1,7);

更简洁的变体

function check(arr, low, high) { 
  for (var i = low; i <= high; i++) {
    var count = arr.filter( d => !(i % d) ).length;
    console.log(i + (count === arr.length ? ' all_match'
                   : count > 0 ? ' one_match'
                   : ''));
  }
};

check([2,3],1,7);