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

Given [3, 8, 4], return 8.

Notes:

1. Single Sequence DP
2. State: f[i] represents the maximal amount which the robber can get at i
3. Function: f[i] = max(f[i-1], f[i-2]+A[i-1])
4. Initial: f[0] = 0; f[1] = A[0];
5. Answer: f[n]
6. Time: O(n), Space: O(n)
7. Optimize: Space -> O(1), rolling array

Space O(n), thinking use this, and then improve

    public long houseRobber1(int[] A) {
        if (A == null || A.length == 0) {
            return 0;
        }

        int n = A.length;
        long[] f = new long[n+1];
        f[0] = 0;
        f[1] = A[0];

        for (int i = 2; i <= A.length; ++i) {
            f[i] = Math.max(f[i-1], f[i-2]+A[i-1]);
        }

        return f[n];
    }

Rolling Array

    public long houseRobber(int[] A) {
        if (A == null || A.length == 0) {
            return 0;
        }

        int n = A.length;
        long[] f = new long[2];
        f[0] = 0;
        f[1] = A[0];

        for (int i = 2; i <= A.length; ++i) {
            f[i%2] = Math.max(f[(i-1)%2], f[(i-2)%2]+A[i-1]);
        }

        return f[n%2];
    }

Two variants

    public long houseRobber2(int[] A) {
        if (A == null || A.length == 0) {
            return 0;
        }

        long pp = 0;
        long p = A[0];
        long current = A[0];

        for (int i = 1; i < A.length; ++i) {
            current = Math.max(p, pp + A[i]);
            pp = p;
            p = current;
        }

        return current;
    }