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leetcode-problemset/leetcode-cn/originData/house-robber-ii.json
2023-12-09 19:57:46 +08:00

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{
"data": {
"question": {
"questionId": "213",
"questionFrontendId": "213",
"categoryTitle": "Algorithms",
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"title": "House Robber II",
"titleSlug": "house-robber-ii",
"content": "<p>You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are <strong>arranged in a circle.</strong> That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and&nbsp;<b>it will automatically contact the police if two adjacent houses were broken into on the same night</b>.</p>\n\n<p>Given an integer array <code>nums</code> representing the amount of money of each house, return <em>the maximum amount of money you can rob tonight <strong>without alerting the police</strong></em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,3,2]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3,1]\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> Rob house 1 (money = 1) and then rob house 3 (money = 3).\nTotal amount you can rob = 1 + 3 = 4.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3]\n<strong>Output:</strong> 3\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 100</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>\n</ul>\n",
"translatedTitle": "打家劫舍 II",
"translatedContent": "<p>你是一个专业的小偷,计划偷窃沿街的房屋,每间房内都藏有一定的现金。这个地方所有的房屋都 <strong>围成一圈</strong> ,这意味着第一个房屋和最后一个房屋是紧挨着的。同时,相邻的房屋装有相互连通的防盗系统,<strong>如果两间相邻的房屋在同一晚上被小偷闯入,系统会自动报警</strong> 。</p>\n\n<p>给定一个代表每个房屋存放金额的非负整数数组,计算你 <strong>在不触动警报装置的情况下</strong> ,今晚能够偷窃到的最高金额。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例&nbsp;1</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = [2,3,2]\n<strong>输出:</strong>3\n<strong>解释:</strong>你不能先偷窃 1 号房屋(金额 = 2然后偷窃 3 号房屋(金额 = 2, 因为他们是相邻的。\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = [1,2,3,1]\n<strong>输出:</strong>4\n<strong>解释:</strong>你可以先偷窃 1 号房屋(金额 = 1然后偷窃 3 号房屋(金额 = 3。\n&nbsp; 偷窃到的最高金额 = 1 + 3 = 4 。</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = [1,2,3]\n<strong>输出:</strong>3\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 100</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>\n</ul>\n",
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"Since House[1] and House[n] are adjacent, they cannot be robbed together. Therefore, the problem becomes to rob either House[1]-House[n-1] or House[2]-House[n], depending on which choice offers more money. Now the problem has degenerated to the <a href =\"https://leetcode.com/problems/house-robber/description/\">House Robber</a>, which is already been solved."
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