如何遍历一个数字数组以查看哪些数字与另一个数字相关？

Question

我正在尝试用原生 JavaScript 编写一个简单的举重程序。用户输入一定量的重量，它会返回特定的重量板放在杠铃的每一侧。

然后我将这个数字带入一个函数，从它减去 45 来计算杠铃重量，然后将该数字除以 2，即放置在杠铃每一侧的重量。

const num = document.getElementById("weightAmount").value;

function getWeightAmount (num) {
const newNum = num - 45;
const halfNum = newNum / 2;
return getWeights(halfNum);
}

每个重量板都有一个数组：

let plates = [44, 33, 22, 11, 5.5, 2.75];

我无法正确循环遍历数组以获取我想要的内容。如果我需要每边 60.5 磅，它应该返回 44、11、5.5。所以我需要弄清楚plate 数组中的哪些数字适合我的第一个函数返回的数字。

我有一个名为 weights 的空数组，我想将 plates 数组中的数字推送到该工作中，然后返回权重。

我的问题是如何循环遍历plates 数组以确定需要哪些权重？

Answer 1

一个可能的解决方案是无限期地迭代，直到任一

• 您有解决方案
• 给定一组权重，问题变得无法解决

每个迭代步骤，您减去可能的最高权重乘以可能的最高因子，将两者存储在合适的数据结构中（我的实现只是使用Object）并继续。

const plates = [44, 33, 22, 11, 5.5, 2.75];

// We assume that plates is always sorted
const determineWeights = (totalWeight) => {
  let factor = 0;
  let weights = {};

  while (totalWeight > 0) {
    weight = plates.find(weight => Math.floor(totalWeight / weight) > 0);

    // There is no weight we can subtract from the total weight to solve the problem
    // Hence, the problem is unsolvable and we return null to indicate that no solution exists
    if (!weight) { return null; }

    // Determine the factor with which to multiply the weight before we subtract from the total weight an subtract the product
    factor = Math.floor(totalWeight / weight);
    totalWeight = totalWeight - factor * weight;

    // Store weight and factor
    weights[weight] = factor;
  }
  
  return weights;
}


console.log(determineWeights(104.5)); // { "11": 1, "44": 2, "5.5": 1 }
console.log(determineWeights(60.5)); // { "11": 1, "44": 1, "5.5": 1 }
console.log(determineWeights(5.0)); // null

问题本质上是Knapsack problem 的一个实例。

请注意，我们假设 plates 已排序。否则，Array.find 不一定会检索到可以从总重量中减去的最大重量。

Answer 2

如果目标重量的值始终是可用盘子的总和，我有一个简单的解决方案。假设权重数组按降序排序。我循环考虑了所有可用的重量，并且只有在总重量超过您需要的总重量时才会继续下一个重量。

function getWeights(targeWeight) {

    let plates = [44, 33, 22, 11, 5.5, 2.75];

    let totalWeight = 0;

    let neededPlates = [];

    let i = 0;

    while(i < plates.length){

        var pweight = totalWeight + plates[i];

        if (pweight > targeWeight) {
            i++;
            continue;
        }

        totalWeight += plates[i];
        neededPlates.push(plates[i]);
    }

    return neededPlates;
}

console.log(getWeights(60.5)); // [44, 11, 5.5]
console.log(getWeights(104.5)); //[44, 44, 11, 5.5]

Answer 3

这里有一个解决方案。如果可用板加起来不等于目标重量，它将返回加起来最接近目标的可用板的组合。改编自this answer。

function createSubsets(numbers, target) {
    // filter out all items larger than target
    numbers = numbers.filter(function (value) {
        return value <= target;
    });

    // sort from largest to smallest
    numbers.sort(function (a, b) {
        return b - a;
    });

    var i;
    var sum = 0;
    var addedIndices = [];

    // go from the largest to the smallest number and
    // add as many of them as long as the sum isn't above target
    for (i = 0; i < numbers.length; i++) {
        if (sum + numbers[i] <= target) {
            sum += numbers[i];
            addedIndices.push(i);
        }
    }

    return addedIndices.map(n => numbers[n]);
}