213. House Robber II

213. House Robber II

题目：

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.

代码如下：

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

解题思路：

这个问题可以简单地分解成两个房屋抢劫者的问题。
假设有n个房子，因为0号住宅和N - 1是邻居，我们不能一起抢劫，因此现在解决方案是最大的。
a.抢劫房屋0至N - 2；
b.抢劫房屋1至N - 1。