## House Robber II

Note:This is an extension of House Robber.After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are

arranged in a circle.That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight

without alerting the police.

Credits:

Special thanks to @Freezen for adding this problem and creating all test cases.

It’s a very similar question to House Robber I, except the houses are circled. The first house and the last house are considered as adjacent.

Find a start point, say nums[0]. Consider two situations, rob it or not. Discuss them separately, and find the maximum result.

//Just decide to rob the first room or not, //divide it into two situations. class Solution { public: int rob(vector<int>& nums) { if(nums.size() == 0) return 0; if(nums.size() == 1) return nums[0]; if(nums.size() == 2) return max(nums[0], nums[1]); //n == 3? //case 1 rob nums[0] int dp[nums.size()]; memset(dp, 0, sizeof(dp)); dp[0] = nums[0]; dp[1] = 0; for(size_t i = 2; i < nums.size(); i++){ for(size_t j = 0; j < i - 1; j++){ int value = nums[i] + dp[j]; if(dp[i] < value){ dp[i] = value; } } } int ans1 = max(dp[nums.size() - 2], dp[nums.size() - 3]); // do not rob the last house memset(dp, 0, sizeof(dp)); dp[0] = 0; dp[1] = nums[1]; for(size_t i = 2; i < nums.size(); i++){ for(size_t j = 0; j < i - 1; j++){ int value = nums[i] + dp[j]; if(dp[i] < value){ dp[i] = value; } } } int ans2 = max(dp[nums.size() - 1], dp[nums.size() - 2]); return max(ans1, ans2); } };

Again, walk through is very important.