feat: 1222

docs/interview/experiences/13202307handwriting.md

@@ -690,6 +690,12 @@ thousand('123456789');
 var num2 = '123456789.1234';
 num2.replace(/(?!^)(?=(\d{3})+\.)/g, ',');
 // '123,456,789.1234'
+
+function thousand(str) {
+  let ans = [];
+  str = str.split('').reverse();
+  for (let i = 0; i < str.length; i++) {}
+}
 ```
 
 ## 23.实现一个 node 异步函数的 promisify

docs/interview/experiences/15202309handwriting.md

@@ -61,12 +61,12 @@ nav:
  * 41.增加数组原型 group 方法
  * 42.Vue 中的响应式机制是如何实现的？请手写代码来实现数据劫持（数据劫持即数据变化时触发回调函数）的简单示例
  * 43.算法：实现一个函数，将给定的十进制数转换为 36 进制表示
- * 44.迭代/递归的方式实现二叉树的层次遍历
+ * 44.useState hook
  * 45.判断一个二叉树是否对称，即左子树和右子树是否镜像对称
  * 46.给定一组乱序的区间，合并重叠的区间并返回结果。
  * 47.多叉树, 获取每一层的节点之和
- * 48.
- * 49.
+ * 48.按照版本号由小到大排序
+ * 49.【代码题】实现一个拼手气抢红包算法
  * 50.
  */
 ```
@@ -1286,7 +1286,21 @@ console.log(res);
 
 ```
 
-## 44.迭代/递归的方式实现二叉树的层次遍历
+## 44.useState hook
+
+```js
+const useState = (defaultVal) => {
+  const val = useRef(defaultVal);
+  const setVal = (newVal) => {
+    if (typeof newVal === 'function') {
+      val.current = newVal(val.current);
+    } else {
+      val.current = newVal;
+    }
+  };
+  return [val, setVal];
+};
+```
 
 ## 45.判断一个二叉树是否对称，即左子树和右子树是否镜像对称
 
@@ -1465,6 +1479,31 @@ var levelOrderDFS = function (root) {
 };
 ```
 
+## 48.按照版本号由小到大排序
+
+```js
+// 样例输入：versions = ['0.1.1', '2.3.3', '0.302.1', '4.2', '4.3.5', '4.3.4.5']
+// 输出：['0.1.1', '0.302.1', '2.3.3', '4.3.4.5', '4.3.5']
+function sortVersion(nums) {
+  return nums.sort((a, b) => {
+    a = a.split('.');
+    b = b.split('.');
+    let len = Math.max(a.length, b.length);
+    for (let i = 0; i < len; i++) {
+      let A = +a[i] || 0;
+      let B = +b[i] || 0;
+      if (A == B) continue;
+      console.log(A - B, 'ab', A, B);
+      return A - B;
+    }
+    return 0;
+  });
+}
+sortVersion(['0.1.1', '2.3.3', '0.302.1', '4.2', '4.3.5', '4.3.4.5']);
+```
+
+## 49.【代码题】实现一个拼手气抢红包算法
+
 ## 参考
 
 - [字节跳动前端面经（3 年工作经验附详细答案）](https://mp.weixin.qq.com/s/MYfuUSNS7xIAT4RgZIMv0g)

docs/interview/experiences/21leetcodeA.md

@@ -410,6 +410,10 @@ function numIslands(grid) {
 ## [17.合并两个有序数组](https://leetcode.cn/problems/merge-sorted-array/)
 
 ```js
+// 输入：nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
+// 输出：[1,2,2,3,5,6]
+// 解释：需要合并 [1,2,3] 和 [2,5,6] 。
+// 合并结果是 [1,2,2,3,5,6] ，其中斜体加粗标注的为 nums1 中的元素。
 function merge(num1, m, num2, n) {
   for (let i = m; i < m + n; i++) {
     num1[i] = nums2[i - m];
@@ -512,6 +516,36 @@ function water(nums) {
 
 ```
 
+## [33.合并区间](https://leetcode.cn/problems/merge-intervals/)
+
+```js
+// 输入：intervals = [[1,3],[2,6],[8,10],[15,18]]
+// 输出：[[1,6],[8,10],[15,18]]
+// 解释：区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].
+function mergeArr(nums) {
+  nums.sort((a, b) => a[0] - b[0]);
+  let ans = [];
+  let pre = nums[0];
+  for (let i = 1; i < nums.length; i++) {
+    let cur = nums[i];
+    if (cur[0] > pre[1]) {
+      ans.push(pre);
+      pre = cur;
+    } else {
+      pre[1] = Math.max(cur[1], pre[1]);
+    }
+  }
+  ans.push(pre);
+  return ans;
+}
+mergeArr([
+  [1, 3],
+  [2, 6],
+  [8, 10],
+  [15, 18],
+]);
+```
+
 ## [38.最长公共子序列](https://leetcode.cn/problems/longest-common-subsequence/)
 
 ```js
@@ -537,3 +571,9 @@ function generate(n) {
   return res;
 }
 ```
+
+## 50.比较版本号
+
+```js
+
+```

docs/interview/experiences/22leetcodeB.md

@@ -46,6 +46,8 @@ nav:
  * 22.二叉树的第K小的元素
  * 23.将有序数组展开为二叉搜索树
  * 24.背包问题
+ * 25.多个数组交集
+ * 26.删除有序数组中的重复项
  */
 ```
 
@@ -90,6 +92,46 @@ var pathSum = function (root, targetSum) {
 
 ## 3.重排链表
 
+## 4.最长公共前缀
+
+```js
+// 输入：strs = ["flower","flow","flight"]
+// 输出："fl"
+// 横行扫描
+function longFind(strs) {
+  if (!strs.length) return '';
+  let prefix = strs[0];
+  for (let i = 0; i < strs.length; i++) {
+    prefix = findL(prefix, strs[i]);
+    if (prefix.length == 0) break;
+  }
+  return prefix;
+}
+function findL(a, b) {
+  let len = Math.min(a.length, b.length);
+  let index = 0;
+  while (a[index] == b[index] && index < len) {
+    index++;
+  }
+  return a.slice(0, index);
+}
+longFind(['flower', 'flow', 'flight']);
+// 纵向扫描
+function longFindHor(strs) {
+  if (!strs.length) return [];
+  let prefix = strs[0];
+  for (let i = 0; i < prefix.length; i++) {
+    for (let j = 1; j < strs.length; j++) {
+      if (strs[j][i] != prefix[i]) {
+        return prefix.slice(0, i);
+      }
+    }
+  }
+  return prefix;
+}
+longFindHor(['flower', 'flow', 'flight']);
+```
+
 ## 6.阶乘(迭代/递归/缓存)
 
 ```js
@@ -354,3 +396,138 @@ var diameterOfBinaryTree = function (root) {
   return total;
 };
 ```
+
+## 24.背包问题
+
+```js
+function weightBag(weight, value, size) {
+  let len = weight.length;
+  let dp = new Array(len).fill().map(() => new Array(size + 1).fill(0));
+  // 首行设置0
+  for (let i = weight[0]; i <= size; i++) {
+    dp[0][i] = value[0];
+  }
+  // 注意遍历顺序
+  for (let i = 1; i < len; i++) {
+    for (let j = 0; j <= size; j++) {
+      if (weight[i] > j) {
+        dp[i][j] = dp[i - 1][j];
+      } else {
+        dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
+      }
+    }
+  }
+  console.table(dp);
+  return dp[len - 1][size];
+}
+weightBag([1, 3, 4], [15, 20, 30], 6);
+```
+
+## 25.多个数组交集
+
+```js
+/**
+ * 多个数组交集 [[3,1,2,4,5],[1,2,3,4],[3,4,5,6]]
+ */
+// 横向扫描
+function intersect2(nums) {
+  if (!nums.length) return [];
+  let prefix = nums[0];
+  for (let i = 1; i < nums.length; i++) {
+    prefix = findT(prefix, nums[i]);
+    if (prefix.length == 0) break;
+  }
+  return prefix;
+}
+function findT(a, b) {
+  let ans = [];
+  for (let item of a) {
+    let index = b.indexOf(item);
+    if (index > -1) {
+      ans.push(item);
+      b.splice(index, 1);
+    }
+  }
+  return ans;
+}
+intersect2([
+  [3, 1, 2, 4, 5],
+  [1, 2, 3, 4],
+  [3, 4, 5, 6],
+]);
+
+// hash
+function intersect3(nums) {
+  let m = new Map();
+  let ans = [];
+  for (let item of nums) {
+    for (let i of item) {
+      if (!m.has(i)) {
+        m.set(i, 0);
+      }
+      let count = m.get(i);
+      m.set(i, count + 1);
+    }
+  }
+  for (let [key, value] of m.entries()) {
+    if (value == nums.length) {
+      ans.push(key);
+    }
+  }
+  return ans;
+}
+intersect3([
+  [3, 1, 2, 4, 5],
+  [1, 2, 3, 4],
+  [3, 4, 5, 6],
+]);
+```
+
+## [26.删除有序数组中的重复项](https://leetcode.cn/problems/remove-duplicates-from-sorted-array/)
+
+```js
+// 输入：nums = [0,0,1,1,1,2,2,3,3,4]
+// 输出：5, nums = [0,1,2,3,4]
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var removeDuplicates = function (nums) {
+  for (let i = 0; i < nums.length; i++) {
+    if (nums[i] == nums[i + 1]) {
+      const v = nums.splice(i, 1);
+      i--;
+    }
+  }
+  return nums.length;
+};
+removeDuplicates([1, 1, 2]);
+function removeDuplicates2(nums) {
+  if (nums.length == 0) return 0;
+  let slow = 0,
+    fast = 1;
+  while (fast < nums.length) {
+    if (nums[fast] != nums[slow]) {
+      slow = slow + 1;
+      nums[slow] = nums[fast];
+    }
+    fast = fast + 1;
+  }
+  return slow + 1;
+}
+//[1,4,1,6]
+function once(nums) {
+  nums.sort((a. b) => a-b)
+  let ans = []
+  for(let i=0;i<nums.length;i++) {
+    if (nums[i] != nums[i++]) {
+      ans.push(nums[i++])
+    } else {
+      if (ans.length) {
+        ans.pop()
+      }
+    }
+  }
+  return ans
+}
+```