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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
0 to n - 2;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<int>& 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<int>& 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 };
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4732035.html
