Leetcode: 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 arearranged 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.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

Understand the problem:
The problem is similar to the last one. The only difference is we can either 
i. Not rob the last one
ii. Not rob the first one.
and calculate the maximum. 

Code (Java):

public class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        if (nums.length == 1) {
            return nums[0];
        }
        
        int n = nums.length;
        // Rob the first house, not the last one. 
        int[] dp = new int[n + 1];
        dp[0] = 0;
        dp[1] = nums[0];
        
        for (int i = 2; i < n; i++) {
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i - 1]);
        }
        
        dp[n] = dp[n - 1];
        
        // No rob the first, might or may not rob  the last one
        int[] dr = new int[n + 1];
        dr[0] = 0;
        dr[1] = 0;
        
        for (int i = 2; i < n + 1; i++) {
            dr[i] = Math.max(dr[i - 1], dr[i - 2] + nums[i - 1]);
        }
        
        return Math.max(dp[n], dr[n]);
    }
}