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‎python-leetcode/.ipynb_checkpoints/1009. 十进制整数的反码-checkpoint.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"#### 题目描述\n",
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"每个非负整数 N 都有其二进制表示。例如， 5 可以被表示为二进制 \"101\"，11 可以用二进制 \"1011\" 表示，依此类推。注意，除 N = 0 外，任何二进制表示中都不含前导零。\n",
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"\n",
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"二进制的反码表示是将每个 1 改为 0 且每个 0 变为 1。例如，二进制数 \"101\" 的二进制反码为 \"010\"。\n",
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"\n",
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"给定十进制数 N，返回其二进制表示的反码所对应的十进制整数。\n",
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"\n",
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" \n",
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"\n",
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"示例 1：\n",
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"\n",
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"- 输入：5\n",
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"- 输出：2\n",
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"- 解释：5 的二进制表示为 \"101\"，其二进制反码为 \"010\"，也就是十进制中的 2 。\n",
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"\n",
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"#### 思路\n",
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"- 十进制转化为二进制\n",
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"- 二进制遍历转化为反码\n",
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"- 反码结果转化为十进制"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 20,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2019-07-03T04:06:31.572554Z",
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"start_time": "2019-07-03T04:06:31.554224Z"
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}
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},
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"outputs": [
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{
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"data": {
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"5"
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]
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},
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"execution_count": 20,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"class Solution:\n",
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" def bitwiseComplement(self, N: int) -> int:\n",
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" result = 0\n",
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" temp = list(bin(N)[2:][::-1])\n",
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" for i, v in enumerate(temp):\n",
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" if v == '1':\n",
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" temp[i] = '0'\n",
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" else:\n",
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" temp[i] = '1'\n",
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" result += int(temp[i])*(2**i)\n",
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" return result\n",
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" \n",
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"N = 10\n",
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"solution = Solution()\n",
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"solution.bitwiseComplement(N)"
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]
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},
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{
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"end_time": "2019-07-03T04:16:04.067626Z",
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"execution_count": 24,
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"metadata": {},
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}
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],
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"source": [
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"class Solution:\n",
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" def bitwiseComplement(self, N: int) -> int:\n",
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" return int(\"\".join(['0' if i == '1' else '1' for i in bin(N)[2:]]), 2)\n",
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" \n",
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"N = 10\n",
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"solution = Solution()\n",
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"solution.bitwiseComplement(N)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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‎python-leetcode/.ipynb_checkpoints/202. 快乐数-checkpoint.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"#### 题目描述\n",
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"编写一个算法来判断一个数是不是“快乐数”。\n",
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"\n",
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"一个“快乐数”定义为：对于一个正整数，每一次将该数替换为它每个位置上的数字的平方和，然后重复这个过程直到这个数变为 1，也可能是无限循环但始终变不到 1。如果可以变为 1，那么这个数就是快乐数。\n",
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"\n",
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"示例: \n",
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"\n",
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"输入: 19\n",
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"输出: true\n",
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"解释: \n",
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"- 12 + 92 = 82\n",
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"- 82 + 22 = 68\n",
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"- 62 + 82 = 100\n",
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"- 12 + 02 + 02 = 1\n",
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"\n",
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"#### 思路\n",
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"- 注意，非快乐树会进入结果为4的死循环"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 25,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2019-07-03T05:19:27.505676Z",
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"start_time": "2019-07-03T05:19:27.490526Z"
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}
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},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"False"
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]
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},
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"execution_count": 25,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"class Solution:\n",
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" def isHappy(self, n: int) -> bool:\n",
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" result = 0\n",
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" for i in str(n):\n",
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" result += int(i)**2\n",
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" if result == 1:\n",
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" return True\n",
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" elif result == 4:\n",
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" return False\n",
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" else:\n",
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" return self.isHappy(result)\n",
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" \n",
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"n = 6\n",
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"solution = Solution()\n",
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"solution.isHappy(n)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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}
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‎python-leetcode/1009. 十进制整数的反码.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"#### 题目描述\n",
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"每个非负整数 N 都有其二进制表示。例如， 5 可以被表示为二进制 \"101\"，11 可以用二进制 \"1011\" 表示，依此类推。注意，除 N = 0 外，任何二进制表示中都不含前导零。\n",
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"\n",
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"二进制的反码表示是将每个 1 改为 0 且每个 0 变为 1。例如，二进制数 \"101\" 的二进制反码为 \"010\"。\n",
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"\n",
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"给定十进制数 N，返回其二进制表示的反码所对应的十进制整数。\n",
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"\n",
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" \n",
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"\n",
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"示例 1：\n",
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"\n",
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"- 输入：5\n",
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"- 输出：2\n",
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"- 解释：5 的二进制表示为 \"101\"，其二进制反码为 \"010\"，也就是十进制中的 2 。\n",
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"\n",
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"#### 思路\n",
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"- 十进制转化为二进制\n",
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"- 二进制遍历转化为反码\n",
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"- 反码结果转化为十进制"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 20,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2019-07-03T04:06:31.572554Z",
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"start_time": "2019-07-03T04:06:31.554224Z"
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}
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},
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"outputs": [
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{
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"data": {
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"5"
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]
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},
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"execution_count": 20,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"class Solution:\n",
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" def bitwiseComplement(self, N: int) -> int:\n",
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" result = 0\n",
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" temp = list(bin(N)[2:][::-1])\n",
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" for i, v in enumerate(temp):\n",
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" if v == '1':\n",
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" temp[i] = '0'\n",
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" else:\n",
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" temp[i] = '1'\n",
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" result += int(temp[i])*(2**i)\n",
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" return result\n",
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" \n",
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"N = 10\n",
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"solution = Solution()\n",
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"solution.bitwiseComplement(N)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 24,
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"metadata": {
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"ExecuteTime": {
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{
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"class Solution:\n",
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" def bitwiseComplement(self, N: int) -> int:\n",
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" return int(\"\".join(['0' if i == '1' else '1' for i in bin(N)[2:]]), 2)\n",
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" \n",
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"N = 10\n",
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"solution = Solution()\n",
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"solution.bitwiseComplement(N)"
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]
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},
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{
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"cell_type": "code",
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