一次代码重构 JavaScript 图连通性判定

简介

说重构其实就是整理了代码，第一次自己手写写的很丑，然后看了书上写的，虽然和书上的思路不同但是整理后几乎一样漂亮

效果

整体代码如下

class Node {
  AdjNodes = new Set()
  id = ""
  constructor(id) { this.id = id }
  connect(node) {
    this.AdjNodes.add(node)
    node.AdjNodes.add(this)
  }
}
class RandomGraph {
  nodes = new Set()
  constructor(size) {
    for (let i = 0; i < size; i++) {
      this.nodes.add(new Node(i))
    }
    const threshold = 1 / size
    for (const x of this.nodes)
      for (const y of this.nodes)
        if (Math.random() < 1 / size)
          x.connect(y)
  }
  print() {
    for (const i of this.nodes) {
      let ids = [...i.AdjNodes]
        .map(x => x.id)
        .join(',')
      console.log(`${i.id}:${ids}`)
    }
  }
  ifConnected() {
    const visit = new Map()

    function* recur(node) {
      if (visit.has(node)) return
      yield node
      for (const x of node.AdjNodes) {
        yield* recur(x)
      }
    }

    let v = this.nodes[Symbol.iterator]().next().value

    for (const i of recur(v))
      visit.set(i)

    return visit.size === this.nodes.size
  }
}
let g = new RandomGraph(6)
g.print()
console.log(`g.ifConnected:${g.ifConnected()}`);

修改之前的代码：

ifConnected() {
  for (const [v,_] of this.nodes.entries()) {
    let visit = new WeakMap()
    function* recur(node) {
      if (visit.has(node)) return
      yield (()=>{
        visit.set(node)
        return node
      })();
      for(const [x] of node.AdjNodes.entries()) {
        yield* recur(x)
      }
    }
    const a = new Set()
    for (const i of recur(v)) {
      console.log(i)
      a.add(i)
    }
    console.log(`node:${v.id} minG.size:${a.size}`)
    if (a.size === this.nodes.size) return true
    return false
  }
}