0213. House Robber II

https://leetcode.com/problems/house-robber-ii

Description

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 a security system connected, and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

**Input:** nums = [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:** nums = [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.

Example 3:

**Input:** nums = [1,2,3]
**Output:** 3

Constraints:

• 1 <= nums.length <= 100

• 0 <= nums[i] <= 1000

ac1

two pass without a function to wrap it up, a little bit ugly.

// so DP doesn't care about those passed state, can't go back to change it.

class Solution {
    public int rob(int[] nums) {
        // edge case, length <= 3, rob 1
        if (nums.length < 4) {
            int res = 0;
            for (int n : nums) {
                res = Math.max(res, n);
            }
            return res;
        }

        // rob first, walk 0 - [nums.length - 2]
        int prev = 0;
        int max = nums[0];
        for (int i = 1; i < nums.length - 1; i++) {
            int tmp = max;
            max = Math.max(max, nums[i] + prev);
            prev = tmp;
        }

        // don't rob first, walk 1 -> [nums.length - 1]
        int max2 = nums[1];
        prev = 0;
        for (int i = 2; i < nums.length; i++) {
            int tmp = max2;
            max2 = Math.max(max2, nums[i] + prev);
            prev = tmp;
        }

        // answer
        return Math.max(max, max2);
    }
}