leetcode 213. House Robber II

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解决环形排列房屋的抢劫问题，确保不连续抢劫相邻房屋，采用动态规划算法找到最大抢劫金额。

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题目描述：

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.

关键就是最后那一个房间N和第一个房间相连了，可以这么进行转化：

考虑抢劫了第0个房间，那么问题就是求抢劫第0~N-1个房间的最大数。

考虑不抢劫第0个房间，那么问题就是求抢劫第1~N个房间的最大数。

实现：

class Solution {
    public int rob(int[] nums) {
        if(nums.length==0) return 0;
        if(nums.length==1) return nums[0];
        if(nums.length==2) return Math.max(nums[0],nums[1]);
        return Math.max(robb(nums,0,nums.length-2),robb(nums,1,nums.length-1));
        
    }
    private int robb(int[] nums,int start,int end){
        int n=end-start+1;
		int[] dp=new int[n];
		dp[0]=nums[start];
		dp[1]=Math.max(nums[start], nums[start+1]);
		for(int i=2;i<n;i++){
			dp[i]=Math.max(dp[i-2]+nums[start+i], dp[i-1]);
		}	
		return dp[n-1];
    }
}