Project Notes

House Robber II

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Description

tab: English
 
<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>
 
<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>
 
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
 
<pre>
<strong>Input:</strong> nums = [2,3,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong> You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.
</pre>
 
<p><strong class="example">Example 2:</strong></p>
 
<pre>
<strong>Input:</strong> nums = [1,2,3,1]
<strong>Output:</strong> 4
<strong>Explanation:</strong> Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.
</pre>
 
<p><strong class="example">Example 3:</strong></p>
 
<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> 3
</pre>
 
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
 
<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
</ul>
 
 
 
> [!tip]- Hint 1
> 
> 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.
 
 
---
 
[submissions](https://leetcode.com/problems/house-robber-ii/submissions/) | [solutions](https://leetcode.com/problems/house-robber-ii/solutions/)
 
 
tab: 中文
 
<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>
 
<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>
 
<p>&nbsp;</p>
 
<p><strong>示例&nbsp;1：</strong></p>
 
<pre>
<strong>输入：</strong>nums = [2,3,2]
<strong>输出：</strong>3
<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。
</pre>
 
<p><strong>示例 2：</strong></p>
 
<pre>
<strong>输入：</strong>nums = [1,2,3,1]
<strong>输出：</strong>4
<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）。
&nbsp;    偷窃到的最高金额 = 1 + 3 = 4 。</pre>
 
<p><strong>示例 3：</strong></p>
 
<pre>
<strong>输入：</strong>nums = [1,2,3]
<strong>输出：</strong>3
</pre>
 
<p>&nbsp;</p>
 
<p><strong>提示：</strong></p>
 
<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
</ul>
 
 
 
> [!tip]- 提示 1
> 
> 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.
 
 
---
 
[提交记录](https://leetcode.cn/problems/house-robber-ii/submissions/) | [题解](https://leetcode.cn/problems/house-robber-ii/solution/)

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