Leetcode 213 House Robber II

原题：
You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

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.

Example 1:

Input: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
because they are adjacent houses.
Example 2:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.

中文：
你是个贼，现在让你偷一个条环形排列的房子，要求当前正在作案的房子相邻的房子不能被偷过。现在问你赃款最大可以是多少？

代码：

class Solution {
public:
    int rob(vector<int>& nums) {
        if(nums.size()==1)
            return nums.front();
        if(nums.size()==0)
            return 0;
        int* dp1=new int [nums.size()+1];
        int* dp2=new int [nums.size()+1];
        memset(dp1,0,sizeof(int)*(nums.size()+1));
        memset(dp2,0,sizeof(int)*(nums.size()+1));
        for(int i=0;i<nums.size()-1;i++)//抢第一
        {
            if(i>=2)
                dp1[i]=max(dp1[i-1],dp1[i-2]+nums[i]);
            else
            {
                if(i==0)
                    dp1[i]=nums[i];
                else
                    dp1[i]=max(nums[i],dp1[i-1]);
            }
        }
        for(int i=1;i<nums.size();i++)
        {
            if(i>=3)
                dp2[i]=max(dp2[i-1],dp2[i-2]+nums[i]);
            else
            {
                if(i==1)
                    dp2[i]=nums[i];
                else
                    dp2[i]=max(nums[i],dp2[i-1]);
            }
        }
        return max(dp1[nums.size()-2],dp2[nums.size()-1]);
        delete [] dp1;
        delete [] dp2;
    }
};

解答：

相比House Robber这道题题目，只需要特殊考虑头尾两个房子，那么就单独考虑两种情况，一种是对第一个房子下手，那么就不能碰最后一个房子，另一种是放过第一个房子，那么最后一间房子就有可能要下手。

返回最大的那个结果即可