xiao commit

interview_coding/Microsoft Interview.ipynb

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 微软面试题\n",
+    "https://www.nowcoder.com/discuss/954988"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##### 一面\n",
+    "给一个employ和manage的关系 \\\n",
+    "|employ|manage|\n",
+    "| :----: | :----: |\n",
+    "| 2  | 5 |\n",
+    "| 5  | 6 |\n",
+    "| 1  | 6 |\n",
+    "\n",
+    "请把他按管理层级打印成如下json串的形式 \\\n",
+    "[ \\\n",
+    "{\\\n",
+    "id:6\\\n",
+    "sub:[\\\n",
+    "{\\\n",
+    "id:5\\\n",
+    "sub:[\\\n",
+    "{id:2}\\\n",
+    "]\\\n",
+    "}\\\n",
+    "{id:1}\\\n",
+    "]\\\n",
+    "}\\\n",
+    "]\\"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##### Note: 是一道看似简单，实际简单，但是写起来并不容易的题。"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/html": [
+       "<pre style=\"white-space:pre;overflow-x:auto;line-height:normal;font-family:Menlo,'DejaVu Sans Mono',consolas,'Courier New',monospace\"><span style=\"font-weight: bold\">[</span>\n",
+       "  <span style=\"font-weight: bold\">{</span>\n",
+       "    <span style=\"color: #000080; text-decoration-color: #000080; font-weight: bold\">\"id\"</span>: <span style=\"color: #008080; text-decoration-color: #008080; font-weight: bold\">6</span>,\n",
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+       "          <span style=\"font-weight: bold\">{</span>\n",
+       "            <span style=\"color: #000080; text-decoration-color: #000080; font-weight: bold\">\"id\"</span>: <span style=\"color: #008080; text-decoration-color: #008080; font-weight: bold\">2</span>\n",
+       "          <span style=\"font-weight: bold\">}</span>\n",
+       "        <span style=\"font-weight: bold\">]</span>\n",
+       "      <span style=\"font-weight: bold\">}</span>,\n",
+       "      <span style=\"font-weight: bold\">{</span>\n",
+       "        <span style=\"color: #000080; text-decoration-color: #000080; font-weight: bold\">\"id\"</span>: <span style=\"color: #008080; text-decoration-color: #008080; font-weight: bold\">1</span>\n",
+       "      <span style=\"font-weight: bold\">}</span>\n",
+       "    <span style=\"font-weight: bold\">]</span>\n",
+       "  <span style=\"font-weight: bold\">}</span>\n",
+       "<span style=\"font-weight: bold\">]</span>\n",
+       "</pre>\n"
+      ],
+      "text/plain": [
+       "\u001b[1m[\u001b[0m\n",
+       "  \u001b[1m{\u001b[0m\n",
+       "    \u001b[1;34m\"id\"\u001b[0m: \u001b[1;36m6\u001b[0m,\n",
+       "    \u001b[1;34m\"sub\"\u001b[0m: \u001b[1m[\u001b[0m\n",
+       "      \u001b[1m{\u001b[0m\n",
+       "        \u001b[1;34m\"id\"\u001b[0m: \u001b[1;36m5\u001b[0m,\n",
+       "        \u001b[1;34m\"sub\"\u001b[0m: \u001b[1m[\u001b[0m\n",
+       "          \u001b[1m{\u001b[0m\n",
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+       "      \u001b[1m{\u001b[0m\n",
+       "        \u001b[1;34m\"id\"\u001b[0m: \u001b[1;36m1\u001b[0m\n",
+       "      \u001b[1m}\u001b[0m\n",
+       "    \u001b[1m]\u001b[0m\n",
+       "  \u001b[1m}\u001b[0m\n",
+       "\u001b[1m]\u001b[0m\n"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "from typing import Tuple, List, Dict, Any, Set\n",
+    "import rich, json\n",
+    "\n",
+    "ids: Dict[int, List[int]] = {}\n",
+    "\n",
+    "def gen_dict(root: int) -> Dict[Any, Any]:\n",
+    "    sub: List[Dict[Any, Any]] = []\n",
+    "    for i in ids[root]:\n",
+    "        sub.append(gen_dict(i))\n",
+    "    ret: Dict[Any, Any] = {\n",
+    "        \"id\": root\n",
+    "    }\n",
+    "    if sub:\n",
+    "        ret[\"sub\"] = sub\n",
+    "    return ret\n",
+    "\n",
+    "def main(rel: List[Tuple[int, int]]):\n",
+    "    root: Set[int] = set()\n",
+    "    for employ, manager in rel:\n",
+    "        root.add(manager)\n",
+    "        root.discard(employ)\n",
+    "        if employ not in ids:\n",
+    "            ids[employ] = []\n",
+    "        if manager not in ids:\n",
+    "            ids[manager] = []\n",
+    "        if employ not in ids[manager]:\n",
+    "            ids[manager].append(employ)\n",
+    "    ret: List[Dict[Any, Any]] = []\n",
+    "    for id in root:\n",
+    "        ret.append(gen_dict(id))\n",
+    "        \n",
+    "    rich.print_json(json.dumps(ret))\n",
+    "\n",
+    "test = [(2, 5), (5, 6), (1, 6)]\n",
+    "\n",
+    "main(test)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3.6.7 64-bit (system)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.6.7"
+  },
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+  "vscode": {
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+   }
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}

leetcode_practice/leetcode1.py

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+import pandas as pd
+
+df = pd.read_json('train.json')

leetcode_practice/动态规划.md

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+# LeetCode 动态规划题型
+
+// reference:
+https://www.bilibili.com/video/BV1xb411e7ww?spm_id_from=333.337.search-card.all.click
+科技公司面试必考题
+中上难度
+必须掌握
+### 1.如何学习动态规划？
+有规律可循，掌握思想，举一反三
+### 2.动态规划题目特点
+* 1.计数
+-- 有多少种方式走到右下角
+-- 有多少种方式选出K个数使得和是Sum
+* 2.求最大最小值
+-- 从左上角走到右下角路径的最大数字和
+-- 最长上升子序列长度
+* 3.求存在性
+-- 取石子游戏，先手是否必胜
+-- 能不能选出K个数使得和是Sum
+
+### 3.动态规划组成部分
+
+#### 1.确定状态
+确定最后一步，即最优策略中的最后一枚硬币coins[k]
+转化成子问题，即最少的硬币拼出更小的面值amout-coins[k]
+
+#### 2.转移方程
+根据子问题定义直接得到
+f[X] = min{f[X-2]+1, f[X-5]+1, f[X-7]+1}
+
+#### 3.初始条件和边界情况
+f[0] = 0
+如果不能组合出amount, f[amount] = inf
+Note: 细心，考虑周全
+
+#### 4.确定遍历顺序
+利用之前的计算结果
+f[0],f[1],f[2]...f[amount]
+### 常见动态规划类型
+* 1.坐标型动态规划（20%）
+* 2.序列型动态规划（20%）
+* 3.划分型动态规划（20%）
+* 4.背包型动态规划
+* 5.最长序列型动态规划
+* 6.博弈型动态规划
+* 7.综合性动态规划
+### 动态规划时间空间优化
+FollowUp常考
+
+### 动态规划打印路径
+
+### 动态规划初探
+#### 坐标型动态规划
+* Unique Path
+
+* Unique Path Ⅱ
+网格中有障碍
+-- 1. 确定状态
+最后一步一定是从左边（i,j-1）或者上边(i-1,j)过来
+状态f[i][j]表示从左上角有多少种方式走到格子（i，j）
+-- 2.转移方程
+坐标型动态规划中数组下标[i][j]即坐标（i,j）
+f[i][j] = f[i][j-1] + f[i-1][j]
+-- 3.初始条件和边界情况
+左上角（0，0）或者右下角（m-1，n-1）有障碍，则无法到达，返回0
+格子（i，j）有障碍，则f[i][j]=0表示无法到达此格子
+
+$F[i][j]=\left\{\begin{matrix}
+                0, & & {if（i,j）有障碍}\\
+                1, & & {i=0 且 j=0}\\
+                f[i-1][j], & & {j=1，第一列} \\
+                f[i][j-1], & & {i=1，第一行} \\
+                f[i-1][j] + f[i][j-1], & & {Others}
+                \end{matrix}\right.$
+
+#### 序列型动态规划
+* Paint House
+-- 1.
+-- 2.
+-- 3. 初始条件和边界情况
+设油漆前i栋房子并且房子i-1是红色、蓝色、绿色的最小花费分别为f[i][0],f[i][1],f[i][2]
+初始条件:
+f[0][0] = f[0][1] = f[0][2] = 0
+即不漆任何房子的花费
+-- 4.
+
+
+#### 划分型动态规划
+
+
+下面是我们用Python代码实现的
+```Python
+class Solution:
+    def coinChange(self, coins, amount) -> int:
+        dp = [float('inf')] * (amount + 1)
+        # 初始条件
+        dp[0] = 0 
+        # 确定遍历顺序           
+        for i in range(1, amount + 1):
+            # 最后一枚硬币coins[j]
+            for j in range(len(coins)):
+                # 边界条件
+                if coins[j] <= i and dp[i-coins[j] != float('inf')]:
+                    dp[i] = min(dp[i], dp[i - coins[j]] + 1)
+
+        if dp[amount] == float('inf'):
+            return -1
+        return dp[amount]
+```