213. House Robber II

213. House Robber II

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.

My Solutions:

和上题不同的地方是有环，所以要考虑两种情况：从第一个开始rob，rob到倒数第二个；从第二个开始rob，rob到倒数第一个

class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int len = nums.length;
        if (len == 1) return nums[0];
        if (len == 2) return Math.max(nums[0], nums[1]);
        return Math.max(helper(nums, 0, len - 2), helper(nums, 1, len - 1));
    }

    private int helper(int[] nums, int s, int e) {
        if (s == e) return nums[s];
        int dp[] = new int[nums.length];
        dp[s] = nums[s];
        dp[s + 1] = Math.max(nums[s + 1], nums[s]);
        for (int i = s + 2; i <= e; i++) {
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
        };

        return dp[e];
    }
}

class Solution {
    public int rob(int[] nums) {
        if (nums.length == 1) return nums[0];
        return Math.max(helper(nums, 0, nums.length - 2), helper(nums, 1, nums.length - 1));
        
    }
    
    public int helper(int[] nums, int lo, int hi) {
        int profit = 0;
        int lastP = 0;
        
        for (int i = lo; i <= hi; i++) {
            int temp = lastP;
            lastP = profit;
            profit = Math.max(profit, nums[i] + temp);
        }
        return profit;
    }
}