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python-practice
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‎python-leetcode/.ipynb_checkpoints/258. 各位相加-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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"给定一个非负整数 num，反复将各个位上的数字相加，直到结果为一位数。\n",
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"\n",
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"示例:\n",
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"\n",
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"- 输入: 38\n",
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"- 输出: 2 \n",
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"- 解释: 各位相加的过程为：3 + 8 = 11, 1 + 1 = 2。 由于 2 是一位数，所以返回 2"
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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": 11,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2019-07-02T13:00:51.819778Z",
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"start_time": "2019-07-02T13:00:51.809399Z"
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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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"4"
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]
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},
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"execution_count": 11,
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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(object):\n",
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" def addDigits(self, num):\n",
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" \"\"\"\n",
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" :type num: int\n",
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" :rtype: int\n",
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" \"\"\"\n",
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" result = 0\n",
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" for i in str(num):\n",
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" result += int(i)\n",
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" return result if result <= 9 else self.addDigits(result)\n",
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"\n",
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"num = 22\n",
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"solution = Solution()\n",
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"solution.addDigits(num)"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "python pratice",
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"language": "python",
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"name": "env_name"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.7.1"
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},
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"toc": {
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"base_numbering": 1,
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"nav_menu": {},
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"number_sections": true,
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"sideBar": true,
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"skip_h1_title": false,
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"title_cell": "Table of Contents",
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"title_sidebar": "Contents",
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"toc_cell": false,
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"toc_position": {},
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"toc_section_display": true,
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"toc_window_display": false
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

‎python-leetcode/.ipynb_checkpoints/868. 二进制间距-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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"给定一个正整数 N，找到并返回 N 的二进制表示中两个连续的 1 之间的最长距离。 \n",
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"\n",
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"如果没有两个连续的 1，返回 0 。\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：22\n",
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"- 输出：2\n",
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"\n",
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"解释：\n",
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"- 22 的二进制是 0b10110 。\n",
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"- 在 22 的二进制表示中，有三个 1，组成两对连续的 1 。\n",
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"- 第一对连续的 1 中，两个 1 之间的距离为 2 。\n",
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"- 第二对连续的 1 中，两个 1 之间的距离为 1 。\n",
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"- 答案取两个距离之中最大的，也就是 2 。\n",
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"\n",
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"---\n",
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"#### 思路\n",
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"正常解法\n",
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"- 遍历，记录两个连续的1之间的距离\n",
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"- 比较记录的距离的最大值"
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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": 12,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2019-07-03T03:36:17.949541Z",
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"start_time": "2019-07-03T03:36:17.935767Z"
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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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"2"
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]
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},
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"execution_count": 12,
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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 binaryGap(self, N: int) -> int:\n",
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" result, last_one = 0, 0\n",
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" for i, v in enumerate(bin(N)[2:]):\n",
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" if v == '1':\n",
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" result = max(result, i - last_one)\n",
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" last_one = i\n",
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" return result\n",
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" \n",
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"N = 22\n",
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"solution = Solution()\n",
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"solution.binaryGap(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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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "python pratice",
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"language": "python",
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"name": "env_name"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.7.1"
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},
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"toc": {
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"base_numbering": 1,
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"nav_menu": {},
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"number_sections": true,
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"sideBar": true,
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"skip_h1_title": false,
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"title_cell": "Table of Contents",
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"title_sidebar": "Contents",
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"toc_cell": false,
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"toc_position": {},
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"toc_section_display": true,
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"toc_window_display": false
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

‎python-leetcode/.ipynb_checkpoints/908. 最小差值 I-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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"给定一个整数数组 A，对于每个整数 A[i]，我们可以选择任意 x 满足 -K <= x <= K，并将 x 加到 A[i] 中。\n",
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"\n",
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"在此过程之后，我们得到一些数组 B。\n",
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"\n",
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"返回 B 的最大值和 B 的最小值之间可能存在的最小差值。\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：A = [1], K = 0\n",
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"- 输出：0\n",
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"- 解释：B = [1]\n",
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"\n",
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"示例 2：\n",
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"\n",
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"- 输入：A = [0,10], K = 2\n",
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"- 输出：6\n",
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"- 解释：B = [2,8]"
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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": 1,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2019-07-02T12:10:24.645668Z",
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"start_time": "2019-07-02T12:10:24.632696Z"
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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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"0"
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]
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},
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"execution_count": 1,
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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(object):\n",
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" def smallestRangeI(self, A, K):\n",
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" \"\"\"\n",
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" :type A: List[int]\n",
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" :type K: int\n",
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" :rtype: int\n",
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" \"\"\"\n",
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" return max(0, max(A) - min(A) - 2*K)\n",
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" \n",
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"A = [1]\n",
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"K = 0\n",
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"solution = Solution()\n",
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"solution.smallestRangeI(A, K)"
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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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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "python pratice",
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"language": "python",
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"name": "env_name"
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},
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"language_info": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.7.1"
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},
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"toc": {
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"base_numbering": 1,
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"nav_menu": {},
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"number_sections": true,
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"sideBar": true,
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"skip_h1_title": false,
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"title_cell": "Table of Contents",
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"title_sidebar": "Contents",
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"toc_cell": false,
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"toc_position": {},
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"toc_section_display": true,
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"toc_window_display": false
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

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