在javascript递归函数中增加参数答案

【问题标题】：Increment a parameter in a javascript recursive function在javascript递归函数中增加参数
【发布时间】：2019-12-26 14:09:11
【问题描述】：

我将以下数据存储在一个变量中：

let categories = [
    {
    name: "a",
    selected: false,
    nodes: [
        {
        name: "aa",
        selected: false
      },
      {
        name: "ab",
        selected: true
      },
      {
        name: "ac",
        selected: true
      },
      {
        name: "ad",
        selected: false
      }
    ]
  },
  {
    name: "b",
    selected: false,
    nodes: [
        {
        name: "ba",
        selected: false
      },
      {
        name: "bb",
        selected: true
      },
      {
        name: "bc",
        selected: true
      },
      {
        name: "bd",
        selected: false
      }
    ]
  }
];

我想数一下selected = true有多少物品。
所以我创建了以下函数：

function getSelected(categories, counter = 0) {
    for (let index = 0; index < categories.length; index++) {
        const category = categories[index];
    if (category.selected) {
        counter++;
    }
    if (category.nodes && category.nodes.length) {
        category.nodes.forEach(cat => getSelected([cat], counter));
    }
    }
  return counter;
}

但它总是返回 0。

A working jsFiddle

【问题讨论】：

标签： javascript arrays object recursion

【解决方案1】：

您可以对节点进行递归减少并计数selected。

const
    countSelected = (s, o) => (o.nodes || []).reduce(countSelected, s + o.selected);

let categories = [{ name: "a", selected: false, nodes: [{ name: "aa", selected: false }, { name: "ab", selected: true }, { name: "ac", selected: true }, { name: "ad", selected: false }] }, { name: "b", selected: false, nodes: [{ name: "ba", selected: false }, { name: "bb", selected: true }, { name: "bc", selected: true }, { name: "bd", selected: false }] }],
    count = categories.reduce(countSelected, 0);

console.log(count);

【讨论】：

PS：节点的深度是未知的，所以如果我们有更深的节点，reduce 将无法工作。
它适用于所有嵌套的nodes 属性，因为它使用递归。请看一下countSelected的减少。

【解决方案2】：

这是因为 Number (and other primitives) 函数的参数重复，它创建了一个用参数值初始化的全新变量。

您可以做的是使用函数的返回值并将其相加，或者您可以使用对象作为参数，因为它们是通过引用发送的，因此您保持相同的对象：

let categories = [{name: "a",selected: false,nodes: [{name: "aa",selected: false},{name: "ab",selected: true},{name: "ac",selected: true},{name: "ad",selected: false}]},{name: "b",selected: false,nodes: [{name: "ba",selected: false},{name: "bb",selected: true},{name: "bc",selected: true},{name: "bd",selected: false}]}]

function getSelectedWithObject(categories, counter = {val: 0}) {
    for (let index = 0; index < categories.length; index++) {
        const category = categories[index]
        if (category.selected) {
            counter.val++
        }
        if (category.nodes && category.nodes.length) {
            category.nodes.forEach(cat => getSelectedWithObject([cat], counter))
        }
    }
    return counter.val
}

function getSelectedWithReturnValue(categories) {
    let counter = 0
    for (let index = 0; index < categories.length; index++) {
        const category = categories[index]
        if (category.selected) {
            counter++
        }
        if (category.nodes && category.nodes.length) {
            category.nodes.forEach(cat => counter += getSelectedWithReturnValue([cat]))
        }
    }
    return counter
}

console.log(getSelectedWithObject(categories))
console.log(getSelectedWithReturnValue(categories))

【讨论】：

【解决方案3】：

counter 是一个整数，所以这里是按值传递的。使counter 成为您想要从函数中返回的结果，然后就可以了：

function getSelected(categories) {
    var counter = 0;
    for (let index = 0; index < categories.length; index++) {
        const category = categories[index];
        if (category.selected) {
           counter++;
        }
        if (category.nodes && category.nodes.length) {
           category.nodes.forEach(cat => counter += getSelected([cat]));
        }
    }
    return counter;
}

console.log(getSelected(categories));

【讨论】：