【leetcode】House Robber II

House Robber II

Note: This is an extension of House Robber.

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.

思路：

在robber问题1的基础下，分两种情况讨论就行，一个是第一个房子被抢，另一个没有被抢，然后注意每种情况DP的范围即可。

class Solution {
public:
int rob(vector<int>& nums)
{
int n=nums.size();
if(n==0) return 0;
if(n==1) return nums[0];

int sum1=0;
int res1=0,res2=0;
for(int i=1;i<n;i++)
{
int temp=res1;
res1=max(res1,res2);//记录没有选择房子的最大值；
res2=temp+nums[i]; //记录选择了当前房子的情况
}
sum1=max(res1,res2);

int sum2=nums[0];
res1=0,res2=0;
for(int i=2;i<n-1;i++)
{
int temp=res1;
res1=max(res1,res2);//记录没有选择房子的最大值；
res2=temp+nums[i]; //记录选择了当前房子的情况
}
sum2+=max(res1,res2);

return max(sum1,sum2);
}
};