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 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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/house-robber
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思路：

1. 当 n == 0 时，return 0

2. 当 n == 1 时，return nums[0]

3.当 n >= 2 时，创建数组 dp[n]表示从0~n-1家能够抢到的最大值

dp[0] = nums[0];

dp[1] = max(nums[0], nums[1]);

dp[i] = max(dp[i – 2] + nums[i], dp[i – 1]);  抢这家和上上家，或者不抢这家，抢上一家

4. 最优解为 return dp[n-1]。

class Solution {
public:
    int rob(vector& nums) {
        int size = nums.size();
        if(size == 0) return 0;
        if(size == 1) return nums[0];
        
        vector dp(size, 0);
        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);   
        for(int i = 2; i < size; i++){
            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
        }     
        return dp[size-1];
    }
    
};