算法设计Week10 LeetCode Algorithms Problem #213 House Robber II

题目描述：

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.

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题目分析：

本题是之前House Robber的升级版。题目中表述房间成环形排列，说明了第一个和最后一个房间不能同时抢劫。这样，可以分两次检查房间，第一次检查编号为0到n-2的房间最大总钱数，第二次检查编号为1到n-1的房间最大总钱数，最后两个值中的最大值就是题目要求的结果。
具体代码如下所示，代码的时间复杂度为O(n)，空间复杂度为O(n)。

class Solution {public:    int rob(vector<int>& nums) {        if(nums.size() == 0) return 0;        if(nums.size() == 1) return nums[0];        return max(robs(nums, 0, nums.size() - 2), robs(nums, 1, nums.size() - 1));    }    int robs(vector<int>& nums, int begin, int end){        vector<int> robmoney(end - begin + 1);        robmoney[0] = nums[begin];        robmoney[1] = max(robmoney[0], nums[begin + 1]);        for(int i = 2; i <= end - begin; i++){            robmoney[i] = max(robmoney[i - 1], robmoney[i - 2] + nums[i + begin]);        }        return robmoney[end - begin];    }};