213. House Robber II

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.

public class Solution {
    public int rob(int[] nums) {
        if(nums==null || nums.length==0) return 0;
        if(nums.length==1) return nums[0];//1，2，null这三个edge cases一个都不能少
        if(nums.length==2) return Math.max(nums[0], nums[1]);
        return Math.max(robsub(nums, 0, nums.length-2), robsub(nums, 1, nums.length-1));
    }
    
    private int robsub(int[] nums, int s, int e) {
        int n = e - s + 1;
        int[] d =new int[n];
        d[0] = nums[s];
        d[1] = Math.max(nums[s], nums[s+1]);
        
        for(int i=2; i<n; i++) {
            d[i] = Math.max(d[i-2]+nums[s+i], d[i-1]);
        }
        return d[n-1];
    }
}

由于是circle，所以第一个和最后一个不能同时rob，因此：

分两种情况：1. 选第一个

2.选最后一个（当然到底用不用的上不一定） 

调用的函数直接用house robber I中就可以，最后比较两者取大的。