問題描述
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.
思考，此問題的解跟House Robber有什麼關係
想法
- 顯然，此問題的解不大於House Robber的解
- 如果沒有搶劫最後一個house，則可以搶劫第1個house，如果搶劫了最後一個house，則第一個house不能劫。即無論如何此問題的解都在[0：：n-1]（搶劫了第一個）和[1:：n]（沒有搶劫第一個）中的一個最大的（PS：如果既沒有搶劫第一個也沒有搶劫最後一個，那麼這個解[0：：n-1]和[1:：n]都有包含）。
代碼
class Solution {
public:
int help(vector<int> &a, int start, int end){
int no = 0, yes = 0;
for(int i = start; i < end; i++){
int temp = yes;
yes = max(yes, no + a[i]);
no = max(no, temp);
}
return max(yes,no);
}
int rob(vector<int>& nums) {
int n = nums.size();
if(n == 0)
return 0;
if(n == 1)
return nums[0];
return max(help(nums,0,n - 1), help(nums,1,n));
}
};