House Robber II问题及解法

问题描述：

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.

问题分析：

虽然变成了 一个环状，但实质分析来看，只不过是分成了下面两种情况(0和n-1现在是邻居，不能同时抢)：

①0到n-2

②1到n-1

①和②中能抢到的最大值就是答案

过程详见代码：

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));}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;    }};