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 2:

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.

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

DP

设状态 f[i] 为到位置i时能抢到的金钱最大和，那么状态转移方程如下：

f[i]=max(f[i-1], f[i-2] + nums[i])

其含义是，如果不选择i，则抢到的钱是f[i-1]，如果选择i，则能抢到的钱是f[i-2] + nums[i]。

 

 现在看来，这个题有点像简化版的best time to buy stock with cooldown lol

class Solution {
    public int rob(int[] nums) {
        int n = nums.length;
        if(n == 0) return 0;
        int[] dp = new int[n];
        dp[0] = nums[0];
        
        for(int i = 1; i < n; i++) {
            dp[i] = Math.max(dp[i - 1], (i > 1 ? dp[i - 2] + nums[i] : nums[i]));
        }
        return dp[n - 1];
    }
}