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twotwoba · Apr 2, 2024 · e9a92c5 · e9a92c5
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diff --git a/.vscode/settings.json b/.vscode/settings.json
@@ -0,0 +1,5 @@
+{
+  "cSpell.words": [
+    "adjs"
+  ]
+}
diff --git a/config.toml b/config.toml
@@ -578,7 +578,7 @@ count = 5
 
 #  赞赏配置
 [params.sponsor]
-enable = true
+enable = false
 bio = "如果你觉得这篇文章对你有所帮助，欢迎赞赏~"
 link = "/sponsor"             # 你的赞赏页面的地址
 custom = ""                   # 自定义 HTML

diff --git a/content/posts/algorithm/trick/暴力递归-BFS.md b/content/posts/algorithm/trick/暴力递归-BFS.md
@@ -0,0 +1,167 @@
+---
+title: '暴力递归-BFS'
+date: 2022-10-09T20:49:13+08:00
+lastmod: 2024-03-27
+series: [trick]
+categories: [algorithm]
+---
+
+## 概念
+
+在学习二叉树的时候，层序遍历就是用 BFS 实现的。
+
+从二叉树拓展到多叉树到图，从一个点开始，向四周开始扩散。一般来说，写 BFS 算法都是用「队列」这种数据结构，每次将一个节点周围的所有节点加入队列。
+
+BFS 一大常见用途，就是找 start 到 target 的最近距离，要能把实际问题往这方面转。
+
+```js
+// 二叉树中 start 就是 root
+function bfs(start, target) {
+    const queue = [start]
+    const visited = new Set()
+    visited.add(start)
+
+    let step = 0
+    while (queue.length > 0) {
+        const size = queue.length
+        for (let i = 0; i < size; ++i) {
+            const el = queue.shift()
+            if (el === target) return step // '需要的信息'
+            const adjs = el.adj() // 泛指获取到 el 的所有相邻元素
+            for (let j = 0; j < adjs.length; ++j) {
+                if (!visited.has(adjs[j])) {
+                    queue.push(adjs[j])
+                    visited.push(adjs[j])
+                }
+            }
+        }
+        step++
+    }
+}
+```
+
+## 练习
+
+### lc.111 二叉树的最小深度 easy
+
+```js
+/**
+ * 忍不住上来先来了个 dfs，但不是今天的主角哈😂 ps: 此处用了回溯的思想，也可以用转为子问题的思想
+ * @param {TreeNode} root
+ * @return {number}
+ */
+var minDepth = function (root) {
+    if (root === null) return 0
+    let min = Infinity
+    const dfs = (root, deep) => {
+        if (root == null) return
+        if (root.left == null && root.right == null) {
+            min = Math.min(min, deep)
+            return
+        }
+        deep++
+        dfs(root.left, deep)
+        deep--
+        deep++
+        dfs(root.right, deep)
+        deep--
+    }
+    dfs(root, 1)
+    return min
+}
+```
+
+```js
+/** BFS 版本 */
+/**
+ * @param {TreeNode} root
+ * @return {number}
+ */
+var minDepth = function (root) {
+    // 这里求的就是 root 到 最近叶子结点（target）的距离，明确了这个，剩下的交给手吧~
+    if (root === null) return 0
+    const queue = [root]
+    let deep = 0
+    while (queue.length) {
+        const size = queue.length
+        deep++
+        for (let i = 0; i < size; ++i) {
+            const el = queue.shift()
+            if (el.left === null && el.right === null) return deep
+            if (el.left) queue.push(el.left)
+            if (el.right) queue.push(el.right)
+        }
+    }
+    return deep
+}
+```
+
+不得不说，easy 面前还是有点底气的，😄。
+
+### lc.752 打开转盘锁
+
+这道题是中等题，but，真的有点难度，需要好好分析。
+
+首先毫无疑问，是一个穷举题目，其次，题目一个重点是：**每次旋转都只能旋转一个拨轮的一位数字**，这样我们就能抽象出每个节点的相邻节点了， '0000', '1000','9000','0100','0900'......如是题目就转变成了从 '0000' 到 target 的最短路径问题了。
+
+```js
+/**
+ * @param {string[]} deadends
+ * @param {string} target
+ * @return {number}
+ */
+var openLock = function (deadends, target) {
+    const queue = ['0000']
+    let step = 0
+    const visited = new Set()
+    visited.add('0000')
+    while (queue.length) {
+        const size = queue.length
+        for (let i = 0; i < size; ++i) {
+            const el = queue.shift()
+            if (deadends.includes(el)) continue
+            if (el === target) return step
+            // 把相邻元素加入 queue
+            for (let j = 0; j < 4; j++) {
+                const up = plusOne(el, j)
+                const down = minusOne(el, j)
+                !visited.has(up) && queue.push(up), visited.add(up)
+                !visited.has(down) && queue.push(down), visited.add(down)
+            }
+        }
+        step++
+    }
+    return -1
+}
+// 首先定义 每个转盘 +1，-1 操作
+function plusOne(str, i) {
+    const arr = str.split('')
+    if (arr[i] == '9') {
+        arr[i] = '0'
+    } else {
+        arr[i]++
+    }
+    return arr.join('')
+}
+function minusOne(str, i) {
+    const arr = str.split('')
+    if (arr[i] == '0') {
+        arr[i] = '9'
+    } else {
+        arr[i]--
+    }
+    return arr.join('')
+}
+```
+
+> 提一下 「双向 BFS 优化」：传统的 BFS 框架就是从起点开始向四周扩散，遇到终点时停止；而双向 BFS 则是从起点和终点同时开始扩散，当两边有交集的时候停止。[labuladong](https://labuladong.online/algo/essential-technique/bfs-framework-2/#%E5%9B%9B%E3%80%81%E5%8F%8C%E5%90%91-bfs-%E4%BC%98%E5%8C%96)
+
+### lc.773 滑动谜题
+
+又是一个小时候玩过的经典小游戏。
+
+这道题的难点是，初学者是真的想不到怎么做，知道用什么方法的也在把实际问题转为 BFS 问题上犯了难。一点点分析，1. 每次都是空位置 0 做选择，移动相邻的上下左右的元素到 0 位置。
+
+<!-- ```js
+
+``` -->
diff --git a/content/posts/algorithm/z-others/数组遍历指南.md b/content/posts/algorithm/z-others/数组遍历指南.md
@@ -110,11 +110,11 @@ while (count <= 25) {
 
 ```TS
 // 数组大小为 5 * 5, 斜线对应索引为 [1, 4]，所以外层for遍历斜线
-for (let l = 1; l <= m - 1; ++l) {
+for (let l = 1; l < m; ++l) {
   // 接着分别确定 i 和 j坐标，这个需要找规律（数学归纳法嘛😂）：
   // 1. i 每次是减少 1 的, 区间为 [0，m-1-l]
   // 2. j 不难发现线 l 与坐标 i,j 的关系： j = l + i
-  for (let i = 0; i <= m - 1 - l; ++i) {
+  for (let i = 0; i < m - l; ++i) { // l yueda i de youbianjie yue xiao
     const j = l + i
     console.log(Arr[i][j])
   }

diff --git a/content/posts/english/en.md b/content/posts/english/en.md
@@ -3,6 +3,7 @@ title: '我的英语学习之路'
 date: 2023-01-01T10:30:18+08:00
 lastmod: 2023-10-10
 weight: 1
+draft: true
 ---
 
 > Learning English can make me happy~ -- 2023/01/01
@@ -25,19 +26,19 @@ weight: 1
 
 {{<youtube -UuLXXQn5jA>}}
 
-- 其实那些极个别的词根词缀我认为还是可以记一记的～
-- 另外，个人觉得语境真的也很重要
+-   其实那些极个别的词根词缀我认为还是可以记一记的～
+-   另外，个人觉得语境真的也很重要
 
 #### 推荐
 
-- YouTube 地址：[EnglishClass101](https://www.youtube.com/@EnglishClass101/playlists)
-- 手机 app ：「墨墨背单词」，按照人的记忆曲线背单词，同时提供发音和例句~ 正在坚持打卡中 👊
+-   YouTube 地址：[EnglishClass101](https://www.youtube.com/@EnglishClass101/playlists)
+-   手机 app ：「墨墨背单词」，按照人的记忆曲线背单词，同时提供发音和例句~ 正在坚持打卡中 👊
 
 ### 美式发音：无脑信 Lisa 老师
 
 #### 推荐
 
-- YouTube 地址：[Accurate English](https://www.youtube.com/@AccurateEnglish)
+-   YouTube 地址：[Accurate English](https://www.youtube.com/@AccurateEnglish)
 
 ---
 
@@ -47,7 +48,7 @@ weight: 1
 
 #### 推荐
 
-- [EngVid](https://www.youtube.com/@EngVid/playlists)，这里有初中高级语法完成视频教程，也有雅思托福听说写～。
+-   [EngVid](https://www.youtube.com/@EngVid/playlists)，这里有初中高级语法完成视频教程，也有雅思托福听说写～。
 
 #### 时态小总结
 
@@ -58,15 +59,15 @@ weight: 1
 1. 动词形式：`past & present` 只需要修改 verb 的形态就好了，`feture` 需要在 verb 前加些料：`be going to`、`will`、`might`，就这么简单与 `aspect` 无关，**`but!`**，有时候 `present simple` 或 `present continuous` 也可以表示将来的意思，需要意会哦~ 比如: `what are you doing this weekend / class starts at 10:00`
 2. 分清楚各个 `aspect` 的 `verb forms` 及意义
 
-   - `simple`，描述事实或习惯(重复的 actions/states) `past & verb & will+verb`
-   - `continuous`，描述某个时刻正在做(动作未完成或临时的) `was/were & am/is/are & will be + verb-ing`
-   - `perfect`，潜在意义里连接了两个时间点 `had & have(has) & will have + past`，我的感觉上是到终止日期完成(不一定是个具体的日期)
+    - `simple`，描述事实或习惯(重复的 actions/states) `past & verb & will+verb`
+    - `continuous`，描述某个时刻正在做(动作未完成或临时的) `was/were & am/is/are & will be + verb-ing`
+    - `perfect`，潜在意义里连接了两个时间点 `had & have(has) & will have + past`，我的感觉上是到终止日期完成(不一定是个具体的日期)
 
-     - `By yesterday, I had remembered 54 words` 从过去的一个点到另一个过去的点，译：已经....
-     - `I have remember 60 words now` 从过去一个点到现在，译：已经...
-     - `I will have remember 100 words by Sunday` 从现在到将来的一个点，译：将会/会...
+        - `By yesterday, I had remembered 54 words` 从过去的一个点到另一个过去的点，译：已经....
+        - `I have remember 60 words now` 从过去一个点到现在，译：已经...
+        - `I will have remember 100 words by Sunday` 从现在到将来的一个点，译：将会/会...
 
-   - `perfect continus`，在 perfect 的基础上强调动作未完成或临时的`~ `had & have(has) & will have + been + verb-ing`，有时候可以被 perfect 取代~
+    - `perfect continus`，在 perfect 的基础上强调动作未完成或临时的`~ `had & have(has) & will have + been + verb-ing`，有时候可以被 perfect 取代~
 
 {{< admonition note >}}
 以上是我自己学习后的感悟就记录了下来，其实看文字没啥卵用，重点是一定要理解了
@@ -76,32 +77,32 @@ weight: 1
 
 这三个是比较常用的 `情态动词 + have + 过去分词` 组合形式，需要理解含义，多说。
 
-- `would have`，等价于中文的 `要是...，就/就会...` 的 **`就/就会`**
-  - `If I had studied harder, I would have gone to a better university`
-- `could have`，等价于中文的 `就可以.../本可以.../有可能.../(本)应该.../本来...`
-  - `If I had more time, I could have finished it/ He could have won it if he hadn't...`
-  - `They could have mixed up the date/ I could have lost my keys` _这里表猜测_
-  - `You could have told me sooner` _这里表批评_
-  - `We were really late, and we could have missed our plane, but luckily we just made it` _表示幸运地避免了坏情况_
-- `should have`，等价于中文的 `(本)应该...`, 带有点批评的意思，有时候也表示期望的事情没有发生
-  - `You should have call me like you promised` 表示批评
-  - `They should been here by now` 表示期待的事情未发生
+-   `would have`，等价于中文的 `要是...，就/就会...` 的 **`就/就会`**
+    -   `If I had studied harder, I would have gone to a better university`
+-   `could have`，等价于中文的 `就可以.../本可以.../有可能.../(本)应该.../本来...`
+    -   `If I had more time, I could have finished it/ He could have won it if he hadn't...`
+    -   `They could have mixed up the date/ I could have lost my keys` _这里表猜测_
+    -   `You could have told me sooner` _这里表批评_
+    -   `We were really late, and we could have missed our plane, but luckily we just made it` _表示幸运地避免了坏情况_
+-   `should have`，等价于中文的 `(本)应该...`, 带有点批评的意思，有时候也表示期望的事情没有发生
+    -   `You should have call me like you promised` 表示批评
+    -   `They should been here by now` 表示期待的事情未发生
 
 ---
 
 ## 其他推荐
 
-- YouTuBe 地址：[LearnEnglishWithTVSeries](https://www.youtube.com/@LearnEnglishWithTVSeries/playlists)，看美剧学英语
-- 手机 app：「**voscreen**」，跟随电影配音练听力口语，很方便。PS: 需要有美区 appleID
+-   YouTuBe 地址：[LearnEnglishWithTVSeries](https://www.youtube.com/@LearnEnglishWithTVSeries/playlists)，看美剧学英语
+-   手机 app：「**voscreen**」，跟随电影配音练听力口语，很方便。PS: 需要有美区 appleID
 
 ## 英语杂志推荐
 
-- https://www.economist.com/
-- https://www.vanityfair.com/
-- https://www.rd.com/
-- https://www.newyorker.com/
-- https://www.nationalgeographic.com/
-- https://www.lonelyplanet.com/
+-   https://www.economist.com/
+-   https://www.vanityfair.com/
+-   https://www.rd.com/
+-   https://www.newyorker.com/
+-   https://www.nationalgeographic.com/
+-   https://www.lonelyplanet.com/
 
 ## 最最最重要的
 

diff --git a/content/posts/mybatis/http/TCP-IP网络模型.md → content/posts/http/TCP-IP网络模型.md b/content/posts/mybatis/http/TCP-IP网络模型.md → content/posts/http/TCP-IP网络模型.md
diff --git a/content/posts/mybatis/http/XSS和CSRF.md → content/posts/http/XSS和CSRF.md b/content/posts/mybatis/http/XSS和CSRF.md → content/posts/http/XSS和CSRF.md
diff --git a/content/posts/mybatis/http/http前世今生.md → content/posts/http/http前世今生.md b/content/posts/mybatis/http/http前世今生.md → content/posts/http/http前世今生.md
diff --git a/content/posts/mybatis/http/常见状态码.md → content/posts/http/常见状态码.md b/content/posts/mybatis/http/常见状态码.md → content/posts/http/常见状态码.md
diff --git a/content/posts/mybatis/http/浏览器缓存.md → content/posts/http/浏览器缓存.md b/content/posts/mybatis/http/浏览器缓存.md → content/posts/http/浏览器缓存.md
diff --git a/content/posts/mybatis/http/跨域.md → content/posts/http/跨域.md b/content/posts/mybatis/http/跨域.md → content/posts/http/跨域.md
diff --git a/content/posts/mybatis/http/输入url到渲染.md → content/posts/http/输入url到渲染.md b/content/posts/mybatis/http/输入url到渲染.md → content/posts/http/输入url到渲染.md
diff --git a/content/posts/java/java_1.md → content/posts/java/base/java_1.md b/content/posts/java/java_1.md → content/posts/java/base/java_1.md
@@ -5,6 +5,7 @@ tags: []
 series: [java]
 categories: [study notes]
 weight: 1
+draft: true
 ---
 
 如今只会前端让我感到处处受制于人，很多自己的想法难以实现，严重依赖于后端，为了改变现状，我决定今年另一个小目标就是掌握 java，以及它的相关技术栈，不折腾一下，做什么程序员？Let's go!

diff --git a/content/posts/java/java_10.md → content/posts/java/base/java_10.md b/content/posts/java/java_10.md → content/posts/java/base/java_10.md
@@ -5,6 +5,7 @@ tags: []
 series: [java]
 categories: [study notes]
 weight: 10
+draft: true
 ---
 
 前端学过 TypeScript，对泛型是一定的理解，来看看 java 的泛型吧。

diff --git a/content/posts/java/java_11.md → content/posts/java/base/java_11.md b/content/posts/java/java_11.md → content/posts/java/base/java_11.md
@@ -6,6 +6,7 @@ tags: []
 series: [java]
 categories: [study notes]
 weight: 11
+draft: true
 ---
 
 ## JFrame & JPanel

diff --git a/content/posts/java/java_12.md → content/posts/java/base/java_12.md b/content/posts/java/java_12.md → content/posts/java/base/java_12.md
@@ -5,6 +5,7 @@ lastmod:
 series: [java]
 categories: [study notes]
 weight: 12
+draft: true
 ---
 
 ## IO 流

diff --git a/content/posts/java/java_13.md → content/posts/java/base/java_13.md b/content/posts/java/java_13.md → content/posts/java/base/java_13.md
@@ -5,6 +5,7 @@ lastmod:
 series: [java]
 categories: [study notes]
 weight: 13
+draft: true
 ---
 
 作为前端开发，对网络这块大体还是比较熟悉的，如需要深度学习网络，可以后续深度学习～

diff --git a/content/posts/java/java_14.md → content/posts/java/base/java_14.md b/content/posts/java/java_14.md → content/posts/java/base/java_14.md
@@ -5,6 +5,7 @@ lastmod:
 series: [java]
 categories: [study notes]
 weight: 14
+draft: true
 ---
 
 ## 场景

diff --git a/content/posts/java/java_2.md → content/posts/java/base/java_2.md b/content/posts/java/java_2.md → content/posts/java/base/java_2.md
@@ -5,6 +5,7 @@ tags: []
 series: [java]
 categories: [study notes]
 weight: 2
+draft: true
 ---
 
 ## JAVA 基本数据类型

diff --git a/content/posts/java/java_3.md → content/posts/java/base/java_3.md b/content/posts/java/java_3.md → content/posts/java/base/java_3.md
@@ -5,6 +5,7 @@ tags: []
 series: [java]
 categories: [study notes]
 weight: 3
+draft: true
 ---
 
 面向对象编程三大特性：`封装、继承和多态`。

diff --git a/content/posts/java/java_4.md → content/posts/java/base/java_4.md b/content/posts/java/java_4.md → content/posts/java/base/java_4.md
@@ -5,6 +5,7 @@ tags: []
 series: [java]
 categories: [study notes]
 weight: 4
+draft: true
 ---
 
 ## Object 类

diff --git a/content/posts/java/java_5.md → content/posts/java/base/java_5.md b/content/posts/java/java_5.md → content/posts/java/base/java_5.md
@@ -5,6 +5,7 @@ tags: []
 series: [java]
 categories: [study notes]
 weight: 5
+draft: true
 ---
 
 ## 接口