Easy

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.

动态规划四要素：
状态：f[i]表示到第i个房间时，所抢到的最大价值
方程：f[i] = Math.max(f[i - 2] + nums[i], f[i - 1])
初始化: f[0] = nums[0]; f[1] = Math.max(nums[0], nums[1])
答案: f[nums.length - 1]

class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0){
            return 0;
        }
        if (nums.length == 1){
            return nums[0];
        }
        int[] f = new int[nums.length];
        f[0] = nums[0];
        f[1] = Math.max(nums[0], nums[1]);
        for (int i = 2; i < nums.length; i++){
            f[i] = Math.max(f[i - 2] + nums[i], f[i - 1]);
        }
        return f[nums.length - 1];
    }
}

这道题可以用滚动数组进行优化，可节约空间。当我们从状态方程发现f[i] 只和i - 2, i - 1有关，那么我们可以一开始就初始化一个大小为2的int[] f, 然后状态方程都取模，同时最后的结果也要取模。

class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0){
            return 0;
        }
        if (nums.length == 1){
            return nums[0];
        }
        int[] f = new int[2];
        f[0] = nums[0];
        f[1] = Math.max(nums[0], nums[1]);
        for (int i = 2; i < nums.length; i++){
            f[i % 2] = Math.max(f[(i - 2) % 2] + nums[i], f[(i - 1) % 2]);
        }
        return f[(nums.length - 1) % 2];
    }
}