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.
Since, the houses are surrounded like a circle. If we rob index 0, we can't rob index size - 1.
Take an example for break:
[4, 2, 3, 13, 8, 22]:
if I start robbing 4, the max would be 4 + 13 -> 17.
If I dont start robbing 4, the max would be 2+13+22 -> 37.
I thus separate the robbing into two cases:
1) start robbing at index 0, stop at size - 2;
2) start robbing at index 1, stop at size - 1.
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]); vector<int> maxWithFirst(nums.size(), 0); vector<int> maxWithLast(nums.size(), 0); maxWithFirst[0] = nums[0]; maxWithFirst[1] = max(nums[0], nums[1]); maxWithLast[1] = nums[1]; maxWithLast[2] = max(nums[1], nums[2]); for(int i = 2; i < nums.size() - 1; ++i) { maxWithFirst[i] = max(nums[i] + maxWithFirst[i-2], maxWithFirst[i-1]); } for(int j = 3; j < nums.size(); ++j) maxWithLast[j] = max(nums[j] + maxWithLast[j-2], maxWithLast[j-1]); int n = nums.size(); return max(maxWithFirst[n - 2], maxWithLast[n - 1]); }