# LeetCode – House Robber II (Java)

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.

Analysis

This is an extension of House Robber. There are two cases here 1) 1st element is included and last is not included 2) 1st is not included and last is included. Therefore, we can use the similar dynamic programming approach to scan the array twice and get the larger value.

Java Solution

```public int rob(int[] nums) { if(nums==null || nums.length==0) return 0;   if(nums.length==1) return nums[0];   int max1 = robHelper(nums, 0, nums.length-2); int max2 = robHelper(nums, 1, nums.length-1);   return Math.max(max1, max2); }   public int robHelper(int[] nums, int i, int j){ if(i==j){ return nums[i]; }   int[] dp = new int[nums.length]; dp[i]=nums[i]; dp[i+1]=Math.max(nums[i+1], dp[i]);   for(int k=i+2; k<=j; k++){ dp[k]=Math.max(dp[k-1], dp[k-2]+nums[k]); }   return dp[j]; }```
Category >> Algorithms >> Interview
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• Cl

That’s right!

• Suzy

Thanks for sharing great solution! Although the size of the array could be n instead of n+1.
int[] dp = new int[n];