Leetcode: House Robber

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that 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.

Understand the problem:
It is a classic DP problem. Define dp[n + 1], where dp[i] means the maximum money till first i houses. 
dp[0] = 0;
dp[1] = nums[1];

dp[i] = Math.max(dp[i - 2] + nums[i - 1], dp[i - 1]);

and the final status is dp[n].

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];
        }
        
        if (nums.length == 2) {
            return Math.max(nums[0], nums[1]);
        }
        
        int[] dp = new int[nums.length + 1];
        
        dp[0] = 0;
        dp[1] = nums[0];
        
        for (int i = 2; i <= nums.length; i++) {
            dp[i] = Math.max(nums[i - 1] + dp[i - 2], dp[i - 1]);
        }
        
        return dp[nums.length];
    }
}

A constant space solution:
Since now dp[i] only depends on dp[i - 1] and dp[i - 2], we could just use three variables for the DP problem. 

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[] dp = new int[nums.length + 1];
        
        int dp2 = 0;
        int dp1 = nums[0];
        int max = 0;
        
        for (int i = 2; i <= nums.length; i++) {
            max = Math.max(nums[i - 1] + dp2, dp1);
            dp2 = dp1;
            dp1 = max;
        }
        
        return max;
    }
}