[LeetCode] House Robber II

This problem is a little tricky at first glance. However, if you have finished the House Robber problem, this problem can simply be decomposed into two House Robber problems. Suppose there are n houses, since house 0 and n - 1 are now neighbors, we cannot rob them together and thus the solution is now the maximum of

1. Rob houses 0 to n - 2;
2. Rob houses 1 to n - 1.

The code is as follows (some edge cases are handled explicitly).

    int robHelper(vector<int>& nums, int left, int right){
        if (left == right) return nums[left];
        vector<int> money(right - left + 1, 0);
        money[0] = nums[left];
        money[1] = max(nums[left], nums[left + 1]);
        for (int i = 2; i <= right - left; i++)
            money[i] = max(money[i - 1], money[i - 2] + nums[left + i]);
        return money[money.size() - 1];
    }
    int rob(vector<int>& nums) {
        if (nums.empty()) return 0;
        int n = nums.size();
        if (n == 1) return nums[0];
        return max(robHelper(nums, 0, n - 2), robHelper(nums, 1, n - 1));
    }