如何找到一系列数字的最小公倍数？ [英] How to find the least common multiple of a range of numbers?

问题描述

给定一个两个数字的数组，让他们定义一个数字范围的开始和结束。例如， [2,6] 表示范围2,3,4,5,6。我想编写JavaScript代码来查找范围的最小公倍数。我的代码只适用于小范围，不是像 [1,13] （这是范围1,2,3,4,5,6,7,8 ，9,10,11,12,13），这会导致堆栈溢出。如何有效地找到范围的最小公倍数？

 函数leastCommonMultiple（arr）{
 var minn，最大; 
 if（arr [0]> arr [1]）{
 minn = arr [1]; 
 max = arr [0]; 
} else {
 minn = arr [0]; 
 max = arr [1]; 
} 
函数repeatRecurse（min，max，scm）{
 if（scm％min === 0&& min< max）{
 return repeatRecurse（min +1，max，scm）; 
} else if（scm％min！== 0& min& min< max）{
 return repeatRecurse（minn，max，scm + max）; 
} 
返回scm; 
} 
 return repeatRecurse（minn，max，max）; 
} 
  

解决方案

$ min $ max $ {

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $' var arr = [];
for（var i = min; i <= max; i ++）{
arr.push（i）;
}
return arr;
}

函数gcd（a，b）{
return！b？ a：gcd（b，a％b）;


函数lcm（a，b）{
return（a * b）/ gcd（a，b）;
}

var multiple = min;
range（min，max）.forEach（function（n）{
multiple = lcm（multiple，n）;
}）;

返回倍数;
}

leastCommonMultiple（1，13）; // => 360360

Given an array of two numbers, let them define the start and end of a range of numbers. For example, [2,6] means the range 2,3,4,5,6. I want to write javascript code to find the least common multiple for the range. My code below works for small ranges only, not something like [1,13] (which is the range 1,2,3,4,5,6,7,8,9,10,11,12,13), which causes a stack overflow. How can I efficiently find the least common multiple of a range?

function leastCommonMultiple(arr) {
    var minn, max;
    if ( arr[0] > arr[1] ) {
        minn = arr[1];
        max = arr[0];
    } else {
        minn = arr[0];
        max = arr[1];
    }
    function repeatRecurse(min, max, scm) {
        if ( scm % min === 0 && min < max ) {
            return repeatRecurse(min+1, max, scm);
        } else if ( scm % min !== 0 && min < max ) {
            return repeatRecurse(minn, max, scm+max);
        }
        return scm;
    } 
    return repeatRecurse(minn, max, max);
}

解决方案

I think this gets the job done.

function leastCommonMultiple(min, max) {
    function range(min, max) {
        var arr = [];
        for (var i = min; i <= max; i++) {
            arr.push(i);
        }
        return arr;
    }

    function gcd(a, b) {
        return !b ? a : gcd(b, a % b);
    }

    function lcm(a, b) {
        return (a * b) / gcd(a, b);   
    }

    var multiple = min;
    range(min, max).forEach(function(n) {
        multiple = lcm(multiple, n);
    });

    return multiple;
}

leastCommonMultiple(1, 13); // => 360360

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