LeetCode 213. House Robber II (Medium)

题目描述：

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.

题目大意：有很多栋房子，房子成环形，要偷这些房子里面的钱，不能连续偷两个房子（会被抓），问最多能偷多少钱。

思路：DP。分情况，偷不偷第1栋房，偷就算1到n - 1栋的最大值，不偷就算2到n栋的最大值，求这两个的最大值即可。

c++代码：

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 0)
            return 0;
        else if (nums.size() == 1)
            return nums[0];
        else
            return max(myRob(nums, 0, nums.size() - 1), myRob(nums, 1, nums.size()));
    }
    int myRob(vector<int>& nums, int l, int r)
    {
        int prev = 0;
        int pprev = 0;
        for (size_t i = l; i < r; ++i)
        {
            int temp = prev;
            prev = max(nums[i] + pprev, prev);
            pprev = temp;
        }
        return prev;
    }
};