문제를 본 날: 2022년 10월 31일

문제를 푼 날: 2022년 10월 31일

https://leetcode.com/problems/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 systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: nums = [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: nums = [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.

Constraints:

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 400

Solution

점화식을 생각해보자. DP[i] = max(DP[i-1], DP[i-2] + nums[i]) 가 될 것 같다.

두 가지 경우의 수가 있기 때문이다.

1. 현재 집을 선택한다. = DP[i-2] + nums[i]
2. 선택하지 않는다 = DP[i-1]

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