198. House Robber

Leetcode: 198. 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.

Example 1:

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.

Example 1:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1). Total amount you can rob = 2 + 9 + 1 = 12.

Solution 1:

public class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0)
            return 0;
        int[][] dp = new int[nums.length][2];
        dp[0][0] = 0;
        dp[0][1] = nums[0];
        for(int i = 1; i < dp.length; i++){
            //if don't rob, then current will be previous sum (either robbed or non-robbed)
            dp[i][0] = Math.max(dp[i-1][0],dp[i-1][1]); 
            //if rob, then current will be this num + robbed previous house
            dp[i][1] = dp[i-1][0] + nums[i]; 
        }
        return Math.max(dp[dp.length-1][0], dp[dp.length-1][1]);
    }
}

Solution 2:

class Solution {
    public int rob(int[] nums) {
        int prev = 0, curr = 0;
        for (int n : nums) {
            int temp = curr;
            //if rob this house, then use prev+n, prev
            //if don't rob this house, then current will keep the previous amount
            curr = Math.max(prev + n, curr);
            prev = temp;
        }
        return curr;
    }
}