javascript - 使用表法查找最小公倍数

Question

使用 Table Method 找到所提供参数的最小公倍数可以除以两者，也可以除以这些参数之间范围内的所有序号。只有两个参数。对于ex [1,3]，找到1,2,3的lcm。

注意 - 它可能会产生无限循环

function smallestCommons(arr) {

  var nums = [];
  var multiples = [];
  if(arr[0]>arr[1]) {
    var bigger = arr[0];
  } else {
    var bigger = arr[1];
  }

  for(var i=bigger;i>0;i--) {
    nums.push(i);
    console.log(i);
  }console.log(nums + " nums");

  var sums = 0;

  while(sums != nums.length) {
    for(var k=0;k<nums.length;k++) {
      if(nums[k] % 2 === 0) {
        nums[k] = nums[k]/2;
        multiples.push(2); 
      } else if(nums[k] % 3 === 0) {
        nums[k] = nums[k]/3;
        multiples.push(3); 
      }else if(nums[k] % 5 === 0) {
        nums[k] = nums[k]/5;
        multiples.push(5); 
      }else if(nums[k] % 7 === 0) {
        nums[k] = nums[k]/7;
        multiples.push(7); 
      }else if(nums[k] === 1) {
        break;
      }else {
        nums[k] = nums[k]/nums[k];
        multiples.push(nums[k]);
      }
    }
  for(var j = bigger; j>0;j--) {
     sums = sums + nums[j]; 
   }
  }
 var scm = [multiples].reduce(function(a,b){console.log(a*b)}); return scm
}
smallestCommons([1,5]);

Answer 1

我发现这是一个简单的解决方案，效果非常好；

• 遍历所有可能的数字，从下限输入 (var i) 开始
• 对于每个数字，测试每个数字在输入边界(var j)之间(包括输入边界)的整除性
• 如果我满足所有条件，则返回它作为答案，否则将 i 递增 1，然后重试

click here for explanation of ? operator in variable initialization

function smallestCommons(arr) {


  //set variables for upper and lower bounds 
  //incase they aren't entered in ascending order
  var big = arr[0] < arr[1] ? arr[1]:arr[0],
      small = arr[0] < arr[1] ? arr[0]:arr[1],
      i = small;

  //loop through all numbers, note the possibility of an infinite loop
  while(true){


      //test each number for divisibility by by both upper and lower 
      //bounds, as well as by all sequential numbers inbetween
      for(var j = small; j <= big; j++){

        if(i % j === 0){
          if(j===big){
            return i;
          }
        }else {
          break;
        }
      }   
    i++;
  }
}


smallestCommons([1,5]);  //60