House Robber II

刚开始没有思路，看了一下别人的思路，假设只有第一件房子，没有最后一间，以及没有第一间，只有最后一间。分别考虑这两种情况的money，哪个大就选哪个，所以只要利用house rob I的那个函数即可

class Solution {
public:
int rob(vector<int>& nums) {
if (nums.size() <= 1) return nums.empty() ? 0 : nums[0];
return max(rob(nums, 0, nums.size() - 1), rob(nums, 1, nums.size()));
}
int rob(vector<int> &nums, int left, int right) {
if (right - left <= 1) return nums[left];
vector<int> dp(right, 0);
dp[left] = nums[left];
dp[left + 1] = max(nums[left], nums[left + 1]);
for (int i = left + 2; i < right; ++i) {
dp[i] = max(nums[i] + dp[i - 2], dp[i - 1]);
}
return dp.back();
}
};

class Solution {
public:
    int rob(vector<int>& nums) {
        const int N = nums.size();
        if (N == 0) return 0;
        if (N < 3)
            return *max_element(nums.begin(), nums.end());
        
        vector<int> nums_first(nums.begin(), --nums.end());
        vector<int> nums_last(++nums.begin(), nums.end());
        
        return max(rob_one(nums_first), rob_one(nums_last));
    }
    
    int rob_one(vector<int>& nums) 
    {
        const int N = nums.size();
        int preprev = 0;
        int prev = 0;
        for (int i = N - 1; i >= 0; --i)
        {
            const int x = nums[i];
            const int tmp = max(x + preprev, prev);
            preprev = prev;
            prev = tmp;
        }
        return prev;
    }
};