feat: 0322

docs/interview/experiences/15202309handwriting.md

@@ -939,6 +939,7 @@ function isCycleDfs(obj) {
       if (cache.has(val)) {
         return true;
       }
+      // 如果不是引用类型直接跳过
       if (typeof val !== 'object' || val == null) continue;
       cache.add(val);
       let flag = helper(val);

docs/interview/experiences/18middleHandwriting.md

@@ -14,24 +14,113 @@ nav:
 
 ## 1.大数相乘
 
+```js
+function multiple(a, b) {
+  let [m, n] = [a.length, b.length];
+  let res = new Array(m + n).fill(0);
+  for (let i = m - 1; i >= 0; i--) {
+    for (let j = n - 1; j >= 0; j--) {
+      let p1 = i + j;
+      let p2 = p1 + 1;
+      let tmp = Number(a[i]) * Number(b[j]);
+      let data = res[p2] + tmp;
+      res[p2] = data % 10;
+      res[p1] = Math.floor(data / 10) + res[p1];
+    }
+  }
+  while (res[0] === 0) {
+    res.shift();
+  }
+  return res.length ? res.join('') : '0';
+}
+multiple('22', '22');
+```
+
 ## 2.数字转汉字
 
 ## 3.归并排序/堆排序
 
 ## 4.数组转树
 
+```js
+
+```
+
 ## 5.树转数组
 
 ## 6.判断对象是否存在循环引用
 
+```js
+function isCycleObj(obj) {
+  let cache = new Set();
+  let dfs = (obj) => {
+    let vals = Object.values(val);
+    for (val of vals) {
+      if (cache.has(val)) {
+        return true;
+      }
+      if (typeof val != 'object' || val == null) continue;
+      cache.add(val);
+      if (dfs(val)) {
+        return true;
+      }
+    }
+    return false;
+  };
+  return dfs(obj);
+}
+```
+
 ## 7.抢红包算法
 
+```js
+function redPacket(total, num, max = 2, min = '0.1') {
+  function name(params) {
+    let remain = total;
+    let ans = [];
+    for (let i = 0; i < num - 1; i++) {
+      let Max = (remain / num) * max;
+      let cur = Math.floor(Max * Math.random() * 100) / 100;
+      cur = cur < min ? min : cur;
+      ans.push(cur);
+      remain = Math.round((remain - cur) * 100) / 100;
+    }
+    ans.push(remain);
+    return ans;
+  }
+  redPacket(10, 4);
+}
+```
+
 ## 8.封装异步的 fetch，使用 async await 方式来使用
 
 ## 9.查找文章中出现频率最高的单词
 
 ## 10.实现双向数据绑定
 
+## 11.判断两个数组内容是否相同
+
+```js
+function isSameArr(a, b) {
+  if (a.length != b.length) return false;
+  let m = new Map();
+  // a塞到m
+  for (let item of a) {
+    if (m.has(item)) {
+      m.set(item, m.get(item) + 1);
+    } else {
+      m.set(item, 1);
+    }
+  }
+  for (let item of b) {
+    let val = m.get(item);
+    if (val == undefined || val < 1) return false;
+    m.set(item, val - 1);
+  }
+  return true;
+}
+```
+
 ## 链接
 
 - [手写 js](https://juejin.cn/post/6946136940164939813?searchId=2024031814180491A668E2D8A6BD15EEE9#heading-55)

docs/interview/experiences/21leetcodeA.md

@@ -133,10 +133,7 @@ function outPutStr(str) {
 ## 2.反转链表
 
 ```js
-/**
- * @param {ListNode} head
- * @return {ListNode}
- */
+// bfs
 var reverseList = function (head) {
   var pre = null;
   var cur = head;
@@ -148,6 +145,14 @@ var reverseList = function (head) {
   }
   return pre;
 };
+// dfs
+function reverseList(head) {
+  if (head == null || head.next == null) return head;
+  let last = reverseList(head.next);
+  head.next.next = head;
+  head.next = null;
+  return last;
+}
 ```
 
 ## 3.LRU 缓存机制

docs/interview/experiences/22leetcodeB.md

@@ -840,6 +840,19 @@ function noMerge(nums) {
   }
   return count;
 }
+// 右区间
+function noMerge2(arr) {
+  arr.sort((a, b) => a[1] - b[1]);
+  let end = arr[0][1];
+  let count = 1;
+  for (let i = 1; i < arr.length; i++) {
+    if (arr[i][0] >= end) {
+      count++;
+      end = arr[i][1];
+    }
+  }
+  return arr.length - count;
+}
 ```
 
 ## [32.寻找两个正序数组的中位数](https://leetcode.cn/problems/median-of-two-sorted-arrays/description/)

docs/interview/experiences/23leetcodeC.md

@@ -41,6 +41,8 @@ nav:
  * 21.分割回文串
  * 22.划分字母区间
  * 23.多数元素
+ * 24.反转链表II
+ * 25.判断两个数组内容相等
  */
 ```
 
@@ -97,7 +99,38 @@ function removeNode(node) {
 
 ## 6.k 个一组反转链表
 
-> 一级难！！！
+> 一级难！！！还是不能理解
+
+```js
+function reverseLink(head, k) {
+  let dummy = { next: head, val: null };
+  let pre = dummy;
+  let cur = head;
+  let len = 0;
+  while (head) {
+    len++;
+    head = head.next;
+  }
+  let count = Math.floor(len / k);
+  for (let i = 0; i < count; i++) {
+    for (let j = 0; j < k - 1; j++) {
+      // pre-1-2-3-4
+      // pre-2-1-3-4
+      // 记录2
+      let next = cur.next;
+      // 1指向 3
+      cur.next = cur.next.next;
+      // 2指向 1
+      next.next = pre.next;
+      // 头节点指向2
+      pre.next = next;
+    }
+    pre = cur;
+    cur = pre.next;
+  }
+  return dummy.next;
+}
+```
 
 ## [7.计算质数的个数](https://leetcode.cn/problems/count-primes/)
 
@@ -121,6 +154,45 @@ function countPrime(n) {
 }
 ```
 
+## 9.打乱数组
+
+```js
+function shufttle(arr) {
+  for (let i = arr.length - 1; i >= 0; i--) {
+    let tmp = Math.floor(Math.random() * (i + 1));
+    [arr[i], arr[tmp]] = [arr[tmp], arr[i]];
+  }
+  return arr;
+}
+shufttle([1, 2, 3, 4, 5, 6]);
+```
+
+## 10.复原 IP 地址
+
+```js
+// 0000 => [0.0.0.0]
+function reverseIp(s) {
+  let ans = [];
+  let backTrack = (path, start) => {
+    if (path.length == 4 && start == s.length) {
+      ans.push(path.slice().join('.'));
+      return;
+    }
+    for (let i = start; i < s.length; i++) {
+      let cur = s.slice(start, i + 1);
+      if (+cur > 255 || cur.length > 3) continue;
+      if (cur.length > 1 && cur[0] == '0') continue;
+      path.push(cur);
+      backTrack(path, i + 1);
+      path.pop();
+    }
+  };
+  backTrack([], 0);
+  return ans;
+}
+reverseIp('0000');
+```
+
 ## 12.两两交换链表节点
 
 ```js
@@ -153,3 +225,32 @@ var swapPairs = function (head) {
 ## 19.小孩报数问题
 
 有 30 个小孩儿，编号从 1-30，围成一圈依此报数，1、2、3 数到 3 的小孩儿退出这个圈， 然后下一个小孩 重新报数 1、2、3，问最后剩下的那个小孩儿的编号是多少?
+
+## 24.反转链表 II
+
+> 跟 k 个一组链表反转 好容易混淆
+
+```js
+// 1->2->3->4->5 2,4
+// 1->4->3->3->5
+function reverseLink(head, left, right) {
+  let dummy = {
+    next: head,
+    val: null,
+  };
+  let pre = dummy;
+  let i = 1;
+  while (i < left) {
+    pre = pre.next;
+    i++;
+  }
+  let cur = pre.next;
+  for (let i = 0; i < right - left; i++) {
+    let next = cur.next;
+    cur.next = next.next;
+    next.next = pre.next;
+    pre.next = next;
+  }
+  return dummy.next;
+}
+```