[LeetCode] House Robber II

This problem is a little tricky at first glance. However, if you have finished the House Robberproblem, 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 (n <= 2) are handled explicitly.

class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size();
        if (n <= 2) return n ? (n == 1 ? nums[0] : max(nums[0], nums[1])) : 0;
        return max(robber(nums, 0, n - 2), robber(nums, 1, n -1));
    }
private:
    int robber(vector<int>& nums, int l, int r) {
        int pre = 0, cur = 0;
        for (int i = l; i <= r; i++) {
            int temp = max(pre + nums[i], cur);
            pre = cur;
            cur = temp;
        } 
        return cur;
    }
};