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.

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];
}
};