feat: 0307

docs/interview/experiences/21leetcodeA.md

@@ -865,6 +865,24 @@ function inorder(node) {
 
 ## 36.二分查找
 
+```js
+function binarySearch(arr, target) {
+  let left = 0;
+  let right = arr.length - 1;
+  while (left <= right) {
+    let base = left + Math.floor(right - left / 2);
+    if (arr[base] == target) {
+      return base;
+    } else if (target > arr[base]) {
+      left = mid + 1;
+    } else {
+      right = mid - 1;
+    }
+  }
+  return -1;
+}
+```
+
 ## [37.用栈实现队列](https://leetcode.cn/problems/implement-queue-using-stacks/description/)
 
 ```js

docs/interview/experiences/4beforeHandwriting.md

@@ -1596,17 +1596,18 @@ camel('cppNName');
 
 输入: nums = [-1,0,3,5,9,12], target = 9 输出: 4 解释: 9 出现在 nums 中并且下标为 4
 
-```ts
-function search(nums: number[], target: number): number {
-  let max = nums.length - 1;
-  let min = 0;
-  while (max >= min) {
-    if (nums[max] === target) {
-      return max;
-    } else if (target < nums[max]) {
-      max--;
-    } else if (target > nums[min]) {
-      min++;
+```js
+function search(nums, target) {
+  let end = nums.length - 1;
+  let start = 0;
+  while (start <= end) {
+    let base = start + end >= 1;
+    if (arr[base] == target) {
+      return base;
+    } else if (arr[base] > target) {
+      end = mid - 1;
+    } else {
+      start = mid + 1;
     }
   }
   return -1;

docs/interview/experiences/practise/202403/0305.js

@@ -12,12 +12,33 @@
  * 11.二叉树中的最大路径和
  */
 
+/**
+ * 
+输入：amount = 5, coins = [1, 2, 5]
+输出：4
+解释：有四种方式可以凑成总金额：
+5=5
+5=2+2+1
+5=2+1+1+1
+5=1+1+1+1+1
+ */
+var change = function (amount, coins) {
+  let dp = new Array(amount + 1).fill(0);
+  dp[0] = 1;
+  for (let coin of coins) {
+    for (let j = coin; j <= amount; j++) {
+      dp[j] += dp[j - coin];
+    }
+  }
+  return dp[amount];
+};
+
 // 1/2 5总共有几种可能
 function clim(n) {
   let dp = new Array(n + 1).fill(0);
   dp[1] = 1;
   dp[2] = 2;
-  for (let i = 2; i <= n; i++) {
+  for (let i = 3; i <= n; i++) {
     dp[i] = dp[i - 1] + dp[i - 2];
   }
   return dp[n];
@@ -33,9 +54,10 @@ function clim(n) {
  */
 var coinChange = function (coins, amount) {
   let dp = new Array(amount + 1).fill(Infinity);
-  for (let i = 1; i <= amount; i++) {
-    for (let item of coins) {
-      dp[i] += dp[i - (amount - item)];
+  dp[0] = 0;
+  for (let coin of coins) {
+    for (let j = coin; j <= amount; j++) {
+      dp[j] = Math.min(dp[j], dp[j - coin] + 1);
     }
   }
   return dp[amount] == Infinity ? '-1' : dp[amount];

docs/interview/experiences/practise/202403/0306.js

@@ -5,15 +5,52 @@
  * 4.Promise.finally
  * 5.reduce
  * 6.二叉树中的最大路径和
- * 7.对称二叉树
+ * 7.对称二叉树/翻转二叉树
  * 8.最大深度和最小深度
  * 9.重排链表
  * 10.最小路径和
  * 11.树转数组
  * 12.数组转树
  */
 
-function tryAgain(fn, time) {}
+/**
+ * @param {TreeNode} root
+ * @return {TreeNode}
+ */
+var invertTree = function (root) {
+  if (!root) return null;
+  let temp = invertTree(root.left);
+  root.left = invertTree(root.right);
+  root.right = temp;
+  return root;
+};
+
+/**
+ * @param {TreeNode} root
+ * @return {boolean}
+ * 外侧
+ * 里侧
+ */
+var isSymmetric = function (root) {
+  if (root == null) return true;
+  const compare = (left, right) => {
+    // 确定终止条件
+    if ((left == null && right != null) || (right == null && left != null)) {
+      return false;
+    }
+    if (left == null && right == null) {
+      return true;
+    }
+    if (left.val !== right.val) {
+      return false;
+    }
+    // 确定单层逻辑
+    const outSide = compare(left.left, right.right);
+    const inner = compare(left.right, right.left);
+    return outSide && inner;
+  };
+  return compare(root.left, root.right);
+};
 
 class Promise {
   static finally(cb) {

docs/interview/experiences/practise/202403/0307.js

@@ -1,8 +1,107 @@
 /**
- * 1.分饼干
+ * 1.分饼干/分糖果
  * 2.删除有序数组的重复项
  * 3.二叉树直径
  * 4.背包问题
  * 5.括号生成
- * 6.
+ * 6.判断子序列
+ * 7.不同的子序列
+ * 8.编辑距离
+ * 9.平衡二叉树
+ * 10.二叉树的最近公共祖先
+ * 11.二叉树的最大深度和最小深度
+ * 12.重排链表
  */
+
+/**
+ * @param {number} n
+ * @return {string[]}
+ * 输入：n = 3
+输出：["((()))","(()())","(())()","()(())","()()()"]
+ */
+var generateParenthesis = function (n) {
+  let ans = [];
+  let dfs = (n, left, right) => {
+    if (left.length > right.length || right.length > n) return;
+    dfs(n);
+  };
+  dfs(n, '(', ')');
+  return ans;
+};
+
+/**
+ * 
+输入: g = [1,2,3], s = [1,1]
+输出: 1
+你有三个孩子和两块小饼干，3个孩子的胃口值分别是：1,2,3。
+虽然你有两块小饼干，由于他们的尺寸都是1，你只能让胃口值是1的孩子满足。
+所以你应该输出1。
+输入: g = [1,2], s = [1,2,3]
+输出: 2
+你有两个孩子和三块小饼干，2个孩子的胃口值分别是1,2。
+你拥有的饼干数量和尺寸都足以让所有孩子满足。
+所以你应该输出2. 
+ */
+var findContentChildren = function (g, s) {
+  g.sort((a, b) => a - b);
+  s.sort((a, b) => a - b);
+  let count = 0;
+  let len = g.length - 1;
+  for (let i = len; i >= 0; i--) {
+    const last = s.slice(-1);
+    if (last >= g[i]) {
+      count++;
+      s.pop();
+    }
+  }
+  return s;
+};
+var findContentChildren = function (g, s) {
+  let count = 0;
+  let len = g.length - 1;
+  let sLen = s.length - 1;
+  for (let i = len; i >= 0; i--) {
+    if (s[sLen] >= g[i]) {
+      count++;
+      sLen--;
+    }
+  }
+  return s;
+};
+
+// weight 1/3/4
+// value 15/20/30
+function weightBag(values, weights, size) {
+  let len = weights.length;
+  // 包 横向
+  // 物品纵向
+  let dp = new Array(len).fill().map(() => new Array(size + 1).fill(0));
+  // 初始化dp
+  for (let i = weight[0]; i <= size; i++) {
+    dp[0][i] = values[0];
+  }
+  // 先遍历物品在遍历背包
+  for (let i = 1; i < len; i++) {
+    for (let j = 0; j <= size; j++) {
+      // 背包塞不下物品
+      if (j < weights[i]) {
+        dp[i][j] = dp[i - 1][j];
+      } else {
+        // 背包塞得下物品
+        dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i]] + values[i]);
+      }
+    }
+  }
+  return dp[len - 1][size];
+}
+
+import { useEffect, useCallback } from 'react';
+export function useEvent(handler) {
+  let handlerRef = useRef();
+  useEffect(() => {
+    handlerRef.current = handler;
+  });
+  return useCallback((...arg) => {
+    handlerRef.current(...arg);
+  }, []);
+}

docs/interview/experiences/practise/202403/0308.js

@@ -1,3 +1,10 @@
 /**
  * 1.每天都在写，到底记住了多少？离03月20日还剩两周，今天要做的就是反思回想
+ * 2.不同路径
+ * 3.从中序和后序遍历构造二叉树
+ * 4.青蛙跳台阶问题
+ * 5.二叉树的层次遍历
+ * 6.寻找重复数
+ * 7.复原IP地址
+ * 8.检测循环依赖
  */

src/Hooks/useToggle/index.md

@@ -0,0 +1,45 @@
+---
+title: useEvent
+order: 6
+group:
+  title: Hooks
+  order: 1,
+nav:
+  order: 3
+  title: 'react'
+  path: /react
+---
+
+## Demo
+
+我最喜欢的自定义 hook 之一是 useToggle，这是一个友好的助手，其工作方式几乎与 useState 完全相同，但只能在 true 和 false 之间切换状态变量:
+
+<code src="./index.tsx"></code>
+
+## Code
+
+```ts
+import React from 'react';
+import useToggle from './useToggle';
+
+const MyApp = () => {
+  const [isDarkMode, toggleDarkMode] = useToggle(false);
+  return <button onClick={toggleDarkMode}>Toggle color theme{isDarkMode}</button>;
+};
+export default MyApp;
+```
+
+## Source code
+
+```ts
+import { useCallback, useState } from 'react';
+
+const useToggle = (initialValue: boolean) => {
+  const [value, setValue] = useState(initialValue);
+  const toggle = useCallback(() => {
+    setValue((v) => !v);
+  }, []);
+  return [value, toggle];
+};
+export default useToggle;
+```

src/Hooks/useToggle/index.tsx

@@ -0,0 +1,8 @@
+import React from 'react';
+import useToggle from './useToggle';
+
+const MyApp = () => {
+    const [isDarkMode, toggleDarkMode] = useToggle(false);
+    return <button onClick={toggleDarkMode}>Toggle color theme:{isDarkMode}</button>;
+};
+export default MyApp;

src/Hooks/useToggle/useToggle.tsx

@@ -0,0 +1,10 @@
+import { useCallback, useState } from 'react';
+
+const useToggle = (initialValue: boolean) => {
+    const [value, setValue] = useState(initialValue);
+    const toggle = useCallback(() => {
+        setValue((v) => !v);
+    }, []);
+    return [value, toggle];
+};
+export default useToggle;