[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.

Credits:
Special thanks to @ifanchu for adding this problem and creating all test cases. Also thanks to @ts for adding additional test cases.

我想到了动态规划的问题。这题可以看做是简单的DP问题，用A[0]表示没有rob当前house的最大money，A[1]表示rob了当前house的最大money，那么A[0] 等于rob或者没有rob上一次house的最大值
即A[i+1][0] = max(A[i][0], A[i][1]).. 那么rob当前的house，只能等于上次没有rob的+money[i+1], 则A[i+1][1] = A[i][0]+money[i+1].
实际上只需要两个变量保存结果就可以了，不需要用二维数组。

public int rob(int[] nums) {
        int best0 = 0; // 表示没有选择当前houses
        int best1 = 0; // 表示选择了当前houses
        for (int i = 0; i < nums.length; i++) {
            int temp = best0;
            best0 = Math.max(best0, best1); // 没有选择当前houses，那么它等于上次选择了或没选择的最大值
            best1 = temp + nums[i]; // 选择了当前houses，值只能等于上次没选择的+当前houses的money
        }
        return Math.max(best0, best1);
    }

提供另外两种动态的问题，我觉得比较难得理解，特别是第二种：

// ---方法一---
    public long houseRobber(int[] A) {
        // write your code here
        int n = A.length;
        if (n == 0)
            return 0;
        long[] res = new long[n + 1];

        res[0] = 0;
        res[1] = A[0];
        for (int i = 2; i <= n; i++) {
            res[i] = Math.max(res[i - 1], res[i - 2] + A[i - 1]);
        }
        return res[n];
    }

    // ---方法二---
    public long houseRobber1(int[] A) {
        // write your code here
        int n = A.length;
        if (n == 0)
            return 0;
        long[] res = new long[2];

        res[0] = 0;
        res[1] = A[0];
        for (int i = 2; i <= n; i++) {
            res[i % 2] = Math
                    .max(res[(i - 1) % 2], res[(i - 2) % 2] + A[i - 1]);
        }
        return res[n % 2];
    }

还有一种方法是状态压缩，暂时还没有研究。