[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.

 

연속으로 훔치지 않고 최대값 구하기

 

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:

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

풀이

- 난이도 easy

- dp 쉬운 버전

- 현재 시점에서 물건을 훔칠 경우와 훔치지 않을 경우의 최대값을 저장한다.

현재 시점에서 물건을 훔친다 = 이전 시점에서 물건을 훔치지 않는다 + 현재 물건 값

현재 시점에서 훔치지 않는다. = max(이전 시점에서 물건을 훔치지 않는다, 이전 시점에서 물건을 훔친다.)

class Solution {
public:
    int rob(vector<int>& nums) {
        int n=nums.size();
        
        if(n == 0){
            return 0;
        }
        
        if(n==1){
            return nums[0];
        }
        
        int dp[100][2];
        memset(dp, 0, sizeof(dp));
        
        dp[0][0] = nums[0];
        for(int i=1; i<n; i++){
            dp[i][0] = nums[i] + dp[i-1][1];
            dp[i][1] = max(dp[i-1][0], dp[i-1][1]);
        }
        return max(dp[n-1][0], dp[n-1][1]);
    }
};