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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 neighboring to each other, the solution is now the maximum of:
1 int robHelper(vector<int>& nums, int left, int right){ 2 if (left == right) return nums[left]; 3 vector<int> money(right - left + 1, 0); 4 money[0] = nums[left]; 5 money[1] = max(nums[left], nums[left + 1]); 6 for (int i = 2; i <= right - left; i++) 7 money[i] = max(money[i - 1], money[i - 2] + nums[left + i]); 8 return money[money.size() - 1]; 9 } 10 int rob(vector<int>& nums) { 11 if (nums.empty()) return 0; 12 int n = nums.size(); 13 if (n == 1) return nums[0]; 14 return max(robHelper(nums, 0, n - 2), robHelper(nums, 1, n - 1)); 15 }
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4548002.html
