Question

此函数使每个元素自身相乘并输出5040

function multiplyAll(arr) {
    var product = 1;

    for (var a = 0; a < arr.length; a++) {
        for (var b = 0; b < arr[a].length; b++) {
            product *= arr[a][b];
        }
    }
    return product;
}
multiplyAll([[1,2],[3,4],[5,6,7]]);

我想要一个显示每个数组中元素相乘的函数（在这种情况下，我希望函数显示[[2]，[12]，[210]]）。

请帮助

Answer 1

您可以使用.map()和.reduce()来实现所需效果，而不会改变输入：

＆＃13;

function multiplyAll (array) {
  return array.map(
    inner => [inner.reduce(
      (product, factor) => product * factor
    )]
  )
}

console.log(multiplyAll([[1,2],[3,4],[5,6,7]]))

＆＃13;

Answer 2

 function multiplyAll(arr) {
    let products = [];

    for (var a = 0; a < arr.length; a++) {
        let product = 1;
        for (var b = 0; b < arr[a].length; b++) {
            product *= arr[a][b];
        }
        products.push([ product ]);
    }

    return products;
 }

 multiplyAll([[1,2],[3,4],[5,6,7]]);

只需构建一个新数组并将结果推送到它。

Answer 3

签出以下代码段，与原始程序相比，其变化最小：

function multiplyAll(arrs) {
    var product = 1;
    var result = [];

    for (var a = 0; a < arrs.length; a++) {
        for (var b = 0; b < arrs[a].length; b++) {
            product *= arrs[a][b];
        }
        result.push(product);
        product=1;
    }
    return result;
}
var product = multiplyAll([[1,2],[3,4],[5,6,7]]);
console.log(product);

Answer 4

检查此解决方案

function multiplyAll(arr) {
 return arr.map((inner) => {
    let product = 1;
    for (let i = 0; i < inner.length; i++) {
      product *= inner[i];
    }
    return [product];
  })
}
let product = multiplyAll([[1,2],[3,4],[5,6,7]]);
console.log(product);

如何在每个数组中获得乘法结果

4 个答案: