Leetcode 213. House Robber II

213. House Robber II

• Total Accepted: 33705
• Total Submissions: 106420
• Difficulty: Medium

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.

思路：基本想法和Leetcode 198. House Robber类似，就是形成了环。不过可以这么考虑：在房间数>=2时，盗窃第0间或者不盗窃第0间，取2种情况的较大值。

代码：

class Solution {
public:
    int rob(int low,int high,vector<int> nums){
        int pre=0,cur=0,i;
        for(i=low;i<=high;i++){
            cur=max(cur+nums[i],pre);
            swap(cur,pre);
        }
        return pre;
    }
    int rob(vector<int>& nums) {
        int n=nums.size();
        if(n<2) return n?nums[0]:0;
        return max(rob(0,n-2,nums),rob(1,n-1,nums));
    }
};