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.

1 class Solution { 2 public: 3     int rob(vector
     
      & nums) { 4         int n = nums.size(); 5         if (n <= 2) return n ? (n == 1 ? nums[0] : max(nums[0], nums[1])) : 0; 6         return max(robber(nums, 0, n - 2), robber(nums, 1, n -1)); 7     } 8 private: 9     int robber(vector
      
       & nums, int l, int r) {10         int pre = 0, cur = 0;11         for (int i = l; i <= r; i++) {12             int temp = max(pre + nums[i], cur);13             pre = cur;14             cur = temp;15         } 16         return cur;17     }18 };