House Robber II
<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>
Array; Dynamic Programming
class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size();
        if (n == 1) {
            return nums[0];
        }
        return max(robRange(nums, 0, n - 2), robRange(nums, 1, n - 1));
    }

    int robRange(vector<int>& nums, int l, int r) {
        int f = 0, g = 0;
        for (; l <= r; ++l) {
            int ff = max(f, g);
            g = f + nums[l];
            f = ff;
        }
        return max(f, g);
    }
};