LeetCode -- House Robber II
题目描述：
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.
与之前的第一个版本类似，只是这次的数组首尾的马会被看成是相邻的，即不能同时抢第1匹和最后1匹。
思路：
本题依然使用DP来解，只是需要基于第1种解法考虑两种特殊情况即可（抢第1匹放弃最后一匹和放弃第1匹抢最后1匹）：
1. 对第[0,n-1]匹马执行DP ， 得到max1
2. 对第[1,n]匹马执行DP，得到max2
最后返回max1与max2的最大值。
注意，由于是环，因此小于4匹马时，只需要返回数组最大值即可。
实现代码：
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.
与之前的第一个版本类似，只是这次的数组首尾的马会被看成是相邻的，即不能同时抢第1匹和最后1匹。
思路：
本题依然使用DP来解，只是需要基于第1种解法考虑两种特殊情况即可（抢第1匹放弃最后一匹和放弃第1匹抢最后1匹）：
1. 对第[0,n-1]匹马执行DP ， 得到max1
2. 对第[1,n]匹马执行DP，得到max2
最后返回max1与max2的最大值。
注意，由于是环，因此小于4匹马时，只需要返回数组最大值即可。
实现代码：
public int Rob(int[] nums)
{
if(nums == null || nums.Length == 0)
{
return 0;
}
if(nums.Length < 4){
return nums.Max();
}
var list = new List<int>(nums);
var first = list[0];
list.RemoveAt(0);
var max1 = Max(list);
list.Insert(0 , first);
list.RemoveAt(list.Count-1);
var max2 = Max(list);
return Math.Max(max1 , max2);
}
private int Max(IList<int> nums)
{
var len = nums.Count;
var dp = new int[len + 1];
dp[0] = 0;
dp[1] = Math.Max(nums[0], 0);
for(var i = 2;i < len + 1; i++){
dp[i] = Math.Max(dp[i-1], dp[i-2] + nums[i-1]);
}
return dp[len];
}