面试题手写数组方法

前端/1html&js/jscode/array/map.md

@@ -0,0 +1,38 @@
+# 实现数组map方法
+
+原文 <https://juejin.cn/post/6844903986479251464#heading-25>
+
+依照 [ecma262 草案](https://tc39.es/ecma262/#sec-array.prototype.map)，实现的map的规范如下:
+
+![ecma262 Array.map](https://p1-jj.byteimg.com/tos-cn-i-t2oaga2asx/gold-user-assets/2019/11/3/16e311d99e860405~tplv-t2oaga2asx-zoom-in-crop-mark:3024:0:0:0.awebp)
+
+```js
+Array.prototype.map = function(callbackFn, thisArg) {
+  // 处理数组类型异常
+  if (this === null || this === undefined) {
+    throw new TypeError("Cannot read property 'map' of null or undefined");
+  }
+  // 处理回调类型异常
+  if (Object.prototype.toString.call(callbackfn) != "[object Function]") {
+    throw new TypeError(callbackfn + ' is not a function')
+  }
+  // 草案中提到要先转换为对象
+  let O = Object(this);
+  let T = thisArg;
+
+  let len = O.length >>> 0; // 字面意思是指"右移 0 位"，但实际上是把前面的空位用0填充，这里的作用是保证len为数字且为整数。
+  let A = new Array(len);
+  for(let k = 0; k < len; k++) {
+    // 还记得原型链那一节提到的 in 吗？in 表示在原型链查找
+    // 如果用 hasOwnProperty 是有问题的，它只能找私有属性
+    // 如果没有找到就不处理，能有效处理稀疏数组的情况。
+    if (k in O) {
+      let kValue = O[k];
+      // 依次传入this, 当前项，当前索引，整个数组
+      let mappedValue = callbackfn.call(T, KValue, k, O);
+      A[k] = mappedValue;
+    }
+  }
+  return A;
+}
+```

前端/1html&js/jscode/array/push pop.md

@@ -0,0 +1,42 @@
+# 实现数组 push、pop 方法
+
+原文 <https://juejin.cn/post/6844903986479251464#heading-27>
+
+参照 ecma262 草案的规定，关于 push 和 pop 的规范如下图所示:
+
+![push](https://p1-jj.byteimg.com/tos-cn-i-t2oaga2asx/gold-user-assets/2019/11/3/16e311f4fa483cc2~tplv-t2oaga2asx-zoom-in-crop-mark:3024:0:0:0.awebp)
+![pop](https://p1-jj.byteimg.com/tos-cn-i-t2oaga2asx/gold-user-assets/2019/11/3/16e311fa338c2ecb~tplv-t2oaga2asx-zoom-in-crop-mark:3024:0:0:0.awebp)
+
+```js
+Array.prototype.push = function(...items) {
+  let O = Object(this);
+  let len = this.length >>> 0;
+  let argCount = items.length >>> 0;
+  // 2 ** 53 - 1 为JS能表示的最大正整数
+  if (len + argCount > 2 ** 53 - 1) {
+    throw new TypeError("The number of array is over the max value restricted!")
+  }
+  for(let i = 0; i < argCount; i++) {
+    O[len + i] = items[i];
+  }
+  let newLength = len + argCount;
+  O.length = newLength;
+  return newLength;
+}
+```
+
+```js
+Array.prototype.pop = function() {
+  let O = Object(this);
+  let len = this.length >>> 0;
+  if (len === 0) {
+    O.length = 0;
+    return undefined;
+  }
+  len --;
+  let value = O[len];
+  delete O[len];
+  O.length = len;
+  return value;
+}
+```

前端/1html&js/jscode/array/reduce.md

@@ -0,0 +1,49 @@
+# 实现数组reduce方法
+
+原文 <https://juejin.cn/post/6844903986479251464#heading-26>
+
+依照 [ecma262 草案](https://p1-jj.byteimg.com/tos-cn-i-t2oaga2asx/gold-user-assets/2019/11/3/16e311ed2bfa8fad~tplv-t2oaga2asx-zoom-in-crop-mark:3024:0:0:0.awebp)，实现的reduce的规范如下:
+
+其中有几个核心要点:
+
+1、初始值不传怎么处理
+2、回调函数的参数有哪些，返回值如何处理。
+
+```js
+Array.prototype.reduce  = function(callbackfn, initialValue) {
+  // 异常处理，和 map 一样
+  // 处理数组类型异常
+  if (this === null || this === undefined) {
+    throw new TypeError("Cannot read property 'reduce' of null or undefined");
+  }
+  // 处理回调类型异常
+  if (Object.prototype.toString.call(callbackfn) != "[object Function]") {
+    throw new TypeError(callbackfn + ' is not a function')
+  }
+  let O = Object(this);
+  let len = O.length >>> 0;
+  let k = 0;
+  let accumulator = initialValue;
+  if (accumulator === undefined) {
+    for(; k < len ; k++) {
+      // 查找原型链
+      if (k in O) {
+        accumulator = O[k];
+        k++;
+        break;
+      }
+    }
+  }
+  // 表示数组全为空
+  if(k === len && accumulator === undefined) 
+    throw new Error('Each element of the array is empty');
+  console.log(k)
+  for(;k < len; k++) {
+    if (k in O) {
+      // 注意，核心！
+      accumulator = callbackfn.call(undefined, accumulator, O[k], k, O);
+    }
+  }
+  return accumulator;
+}
+```

前端/1html&js/算法/0核心框架/0_BFS.js

@@ -10,13 +10,15 @@
 // 将s[j]向上拨动一次
 function plusOne(str, index) {
     const arr = str.split('')
-    arr[index] = arr[index] == 9 ? 0 : arr[index] - 0 + 1
+    const num = Number(arr[index])
+    arr[index] = num === 9 ? 0 : num + 1
     return arr.join('')
 }
 // 将s[j]向下拨动一次
 function minusOne(str, index) {
     const arr = str.split('')
-    arr[index] = arr[index] == 0 ? 9 : arr[index] - 1
+    const num = Number(arr[index])
+    arr[index] = num === 0 ? 9 : arr[index] - 1
     return arr.join('')
 }
 
@@ -74,7 +76,6 @@ function minusOne(str, index) {
 // 双向BFS Set 740 ms	52.8 MB
 const openLock = function (deadends, target) {
     if (deadends.includes(target)) { return -1 }
-    const deads = new Set(deadends)
     const q = []
     const visited = new Set()
     // 从起点开始启动广度优先搜索
@@ -91,11 +92,13 @@ const openLock = function (deadends, target) {
     while (q.length && q1.length) {
         /* 判断是否到达终点 */
         if (q.some(item => q1.includes(item))) {
-            return step - 0 + step1
+            return step + step1
         }
         if (q.some(item => item === target)) { return step }
-        step = run(deadends, q, visited, step)
-        step1 = run(deadends, q1, visited1, step1)
+        run(deadends, q, visited, step)
+        run(deadends, q1, visited1, step1)
+        step++
+        step1++
     }
     return -1
 }
@@ -120,9 +123,8 @@ function run(deadends, q, visited, step) {
             })
         }
     }
-    return step + 1
 }
 
 console.log(openLock(["0201", "0101", "0102", "1212", "2002"], "0202"), 6)
-// console.log(openLock(["0000"], "8888"), -1)
-// console.log(openLock(["8888"], "0009"), 1)
+console.log(openLock(["0000"], "8888"), -1)
+console.log(openLock(["8888"], "0009"), 1)