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‎python-leetcode/.ipynb_checkpoints/682. 棒球比赛-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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"你现在是棒球比赛记录员。\n",
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"给定一个字符串列表，每个字符串可以是以下四种类型之一：\n",
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"\n",
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"1. 整数（一轮的得分）：直接表示您在本轮中获得的积分数。\n",
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"2. \"+\"（一轮的得分）：表示本轮获得的得分是前两轮有效 回合得分的总和。\n",
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"3. \"D\"（一轮的得分）：表示本轮获得的得分是前一轮有效 回合得分的两倍。\n",
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"4. \"C\"（一个操作，这不是一个回合的分数）：表示您获得的最后一个有效 回合的分数是无效的，应该被移除。\n",
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"\n",
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"每一轮的操作都是永久性的，可能会对前一轮和后一轮产生影响。\n",
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"你需要返回你在所有回合中得分的总和。\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\",\"2\",\"C\",\"D\",\"+\"]\n",
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"\n",
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"输出: 30\n",
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"\n",
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"解释: \n",
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"\n",
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"第1轮：你可以得到5分。总和是：5。\n",
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"\n",
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"第2轮：你可以得到2分。总和是：7。\n",
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"\n",
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"操作1：第2轮的数据无效。总和是：5。\n",
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"\n",
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"第3轮：你可以得到10分（第2轮的数据已被删除）。总数是：15。\n",
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"\n",
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"第4轮：你可以得到5 + 10 = 15分。总数是：30。\n",
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"\n",
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"#### 思路：\n",
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"使用栈的思想：\n",
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"- 如果是整数，压栈\n",
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"- 如果是操作符，按对应的运算处理"
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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": 2,
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"metadata": {
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"ExecuteTime": {
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"end_time": "2019-07-01T01:48:00.107253Z",
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"start_time": "2019-07-01T01:48:00.090827Z"
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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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"30"
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]
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},
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"execution_count": 2,
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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 calPoints(self, ops):\n",
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" \"\"\"\n",
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" :type ops: List[str]\n",
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" :rtype: int\n",
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" \"\"\"\n",
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" stack = []\n",
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" for op in ops:\n",
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" if op in \"CD+\":\n",
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" self.cal(stack, op)\n",
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" else:\n",
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" stack.append(int(op))\n",
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" return sum(stack)\n",
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" \n",
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" def cal(self, stack, op):\n",
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" if op == \"C\":\n",
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" if stack :stack.pop()\n",
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" if op == \"D\":\n",
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" stack.append(stack[-1]*2)\n",
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" if op == \"+\":\n",
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" stack.append(stack[-1]+stack[-2])\n",
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" \n",
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"ops = [\"5\",\"2\",\"C\",\"D\",\"+\"]\n",
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"solution = Solution()\n",
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"solution.calPoints(ops)"
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]
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},
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{
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}

‎python-leetcode/682. 棒球比赛.ipynb‎

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{
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"cells": [
3+
{
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"cell_type": "markdown",
5+
"metadata": {},
6+
"source": [
7+
"#### 题目描述\n",
8+
"你现在是棒球比赛记录员。\n",
9+
"给定一个字符串列表，每个字符串可以是以下四种类型之一：\n",
10+
"\n",
11+
"1. 整数（一轮的得分）：直接表示您在本轮中获得的积分数。\n",
12+
"2. \"+\"（一轮的得分）：表示本轮获得的得分是前两轮有效 回合得分的总和。\n",
13+
"3. \"D\"（一轮的得分）：表示本轮获得的得分是前一轮有效 回合得分的两倍。\n",
14+
"4. \"C\"（一个操作，这不是一个回合的分数）：表示您获得的最后一个有效 回合的分数是无效的，应该被移除。\n",
15+
"\n",
16+
"每一轮的操作都是永久性的，可能会对前一轮和后一轮产生影响。\n",
17+
"你需要返回你在所有回合中得分的总和。\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\",\"2\",\"C\",\"D\",\"+\"]\n",
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"\n",
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"输出: 30\n",
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"\n",
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"解释: \n",
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"\n",
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"第1轮：你可以得到5分。总和是：5。\n",
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"\n",
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"第2轮：你可以得到2分。总和是：7。\n",
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"\n",
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"操作1：第2轮的数据无效。总和是：5。\n",
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"\n",
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"第3轮：你可以得到10分（第2轮的数据已被删除）。总数是：15。\n",
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"\n",
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"第4轮：你可以得到5 + 10 = 15分。总数是：30。\n",
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"\n",
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"#### 思路：\n",
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"使用栈的思想：\n",
40+
"- 如果是整数，压栈\n",
41+
"- 如果是操作符，按对应的运算处理"
42+
]
43+
},
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{
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"cell_type": "code",
46+
"execution_count": 2,
47+
"metadata": {
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"ExecuteTime": {
49+
"end_time": "2019-07-01T01:48:00.107253Z",
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"start_time": "2019-07-01T01:48:00.090827Z"
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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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"30"
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]
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},
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"execution_count": 2,
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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 calPoints(self, ops):\n",
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" \"\"\"\n",
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" :type ops: List[str]\n",
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" :rtype: int\n",
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" \"\"\"\n",
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" stack = []\n",
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" for op in ops:\n",
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" if op in \"CD+\":\n",
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" self.cal(stack, op)\n",
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" else:\n",
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" stack.append(int(op))\n",
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" return sum(stack)\n",
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" \n",
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" def cal(self, stack, op):\n",
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" if op == \"C\":\n",
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" if stack :stack.pop()\n",
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" if op == \"D\":\n",
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" stack.append(stack[-1]*2)\n",
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" if op == \"+\":\n",
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" stack.append(stack[-1]+stack[-2])\n",
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" \n",
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"ops = [\"5\",\"2\",\"C\",\"D\",\"+\"]\n",
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"solution = Solution()\n",
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"solution.calPoints(ops)"
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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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"display_name": "python pratice",
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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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}

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