add: 系列手写题

Author: Eric-art-coder

labuladuo/3.无重复字符的最长子串.js

@@ -0,0 +1,16 @@
+/*
+ * @lc app=leetcode.cn id=3 lang=javascript
+ *
+ * [3] 无重复字符的最长子串
+ */
+
+// @lc code=start
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var lengthOfLongestSubstring = function(s) {
+
+};
+// @lc code=end
+

手写专题/js实现一个带并发限制的异步调度器Scheduler.js

@@ -1,17 +1,40 @@
+const timeout = (order, time) => new Promise(resolve => setTimeout(resolve(), time))
+
 class Scheduler {
   constructor() {
     this.maxCount = 2
+    this.runCount = 0
     this.queue = []
     this.run = []
   }
-
-  add(task) {
-    this.queue.push(task)
+  addTask(timer, task) {
+    this.queue.push(timeout(task, timer).then(() => console.log(task)))
   }
-  schedule() {
-    if(this.run.length < this.count && this.queue.length) {
-      const task = this.queue.shift()
-      const 
+  start() {
+    for (let i = 0; i < this.maxCount; i++) {
+      this.request()
     }
   }
-}
+  request() {
+    if(!this.queue || !this.queue.length || this.runCount >= this.maxCount) {
+      return
+    }
+    this.runCount++
+
+    this.queue.shift().then(() => {
+      this.runCount--
+      this.request()
+    })
+  }
+}
+
+
+
+const scheduler = new Scheduler()
+
+scheduler.addTask(1000, '1');
+scheduler.addTask(500, '2');
+scheduler.addTask(300, '3');
+scheduler.addTask(400, '4');
+
+scheduler.start()

手写专题/实现一个带并发限制的Promise.all.js

@@ -0,0 +1,40 @@
+/**
+ * 并发promise
+ * 
+ * @param {*} array 待执行的函数列表
+ * @param {*} limit 并发限制
+ */
+async function asyncPool(array, limit) {
+  // 存储Promise的执行结果
+  const res = []
+  // 正在执行的promise
+  const executing = []
+
+  for (const item of array) {
+    const p = item
+    res.push(p)
+    if(limit <= array.length) {
+      p.then(() => executing.splice(0, 1))
+      executing.push(p)
+      if(executing.length >= limit) await Promise.race(executing)
+    }
+  }
+
+  return Promise.all(res)
+}
+
+const timeout = (order, time) => new Promise(resolve => setTimeout(resolve(console.log(order)), time))
+
+const arr = [
+  timeout(1, 100),
+  timeout(2, 200),
+  timeout(3, 300),
+  timeout(4, 400),
+  timeout(5, 500),
+  timeout(6, 500),
+  timeout(7, 400),
+  timeout(8, 300),
+  timeout(9, 200),
+  timeout(10, 100),
+]
+asyncPool(arr, 10).then(result => console.log(result))

手写专题/数组转树状结构.js

@@ -0,0 +1,51 @@
+// 使用递归的方式实现
+const data = [
+  { id: '01', name: '张大大', pid: '', job: '项目经理' },
+  { id: '02', name: '小亮', pid: '01', job: '产品leader' },
+  { id: '03', name: '小美', pid: '01', job: 'UIleader' },
+  { id: '04', name: '老马', pid: '01', job: '技术leader' },
+  { id: '05', name: '老王', pid: '01', job: '测试leader' },
+  { id: '06', name: '老李', pid: '01', job: '运维leader' },
+  { id: '07', name: '小丽', pid: '02', job: '产品经理' },
+  { id: '08', name: '大光', pid: '02', job: '产品经理' },
+  { id: '09', name: '小高', pid: '03', job: 'UI设计师' },
+  { id: '10', name: '小刘', pid: '04', job: '前端工程师' },
+  { id: '11', name: '小华', pid: '04', job: '后端工程师' },
+  { id: '12', name: '小李', pid: '04', job: '后端工程师' },
+  { id: '13', name: '小赵', pid: '05', job: '测试工程师' },
+  { id: '14', name: '小强', pid: '05', job: '测试工程师' },
+  { id: '15', name: '小涛', pid: '06', job: '运维工程师' }
+]
+
+// 用reduce实现方式
+// function toTree(arr, parentId){
+// 	function dp(parId) {
+//     return arr.reduce((acc, cur) => {
+//       if(cur.pid === parId) {
+//         cur.children = dp(cur.id)
+//         acc.push(cur)
+//       }
+//       return acc
+//     }, [])
+//   }
+//   return dp(parentId)
+// }
+
+// 用for循环
+function toTree(arr, parentId) {
+  function dp(parId) {
+    let res = []
+    for (let i = 0; i < arr.length; i++) {
+      const item = arr[i];
+      if(item.pid === parId) {
+        item.children = dp(item.id)
+        res.push(item)
+      }
+    }
+    return res
+  }
+  return dp(parentId)
+}
+
+
+console.log(toTree(data, ''))

手写专题/最长不重复字符串.js

@@ -0,0 +1,26 @@
+/**
+输入: s = "abcabcbb"
+输出: 3 
+解释: 因为无重复字符的最长子串是 "abc"，所以其长度为 3。
+ */
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var lengthOfLongestSubstring = function(s) {
+  let myMap = new Map()
+  let l = 0, r = 0, maxl = 0
+  while(r < s.length) {
+    if(myMap.has(s[r])) {
+      let i = myMap.get(s[r]) + 1
+      l = Math.max(l, i)
+    }
+    myMap.set(s[r], r)
+    maxl = Math.max(maxl, r - l + 1)
+    r++
+  }
+  return maxl
+}
+
+var s = "abcabcbb"
+console.log(lengthOfLongestSubstring(s))