House Robber II (leetcode 213)

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.

思路： 在原题基础上 二选一 一种不选第一个，第二种， 不选最后一个。其它按照直线方式。

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 0) {
            return 0;
        } 
        if (nums.size() == 1) {
            return nums[0];
        }
        vector<int> a(nums);
        vector<int> b(nums);
        a.pop_back();
        b.erase(b.begin());
        int aa = rob1(a);
        int bb = rob1(b);
        return aa > bb ? aa : bb;
    }
    int rob1(vector<int>& nums) {
        int n =nums.size();
        if(n == 0)
            return 0;
        else if(n == 1)
            return nums[0];
        else
        {
            vector<int> maxV(n, 0);
            maxV[0] = nums[0];
            maxV[1] = max(nums[0], nums[1]);
            for(int i = 2; i < n; i ++)
                maxV[i] = max(maxV[i-2]+nums[i], maxV[i-1]);
            return maxV[n-1];
        }
    }
};