###### tags: `Leetcode` `medium` `dynamic programming` `python` `c++` `Top 100 Liked Questions`

# 198. House Robber

## [題目連結:] https://leetcode.com/problems/house-robber/description/

## 題目:
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.

## 解題想法:
* 題目為: 小偷要搶錢，但是不能連續搶鄰居，求能搶的最大值
* 透過DP:
* dp[i]表示搶到第i個房子時候
* 轉移方程:
* dp[i]=max(dp[i-2]+nums[i],dp[i-1])
* 因為不能搶鄰居，所以考慮當前nums[i]是否搶時:
* 選擇dp[i-2]+nums[i]
* 或是不搶nums[i]，繼承dp[i-1]
* 對於初始:
* dp[0]=nums[0]
* **dp[1]=max(nums[0],nums[1])**
* 因為不能打劫相鄰的房子，所以搶nums[0] or nums[1]
## Python:

class Solution(object):
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        #time space: O(n)
        #dp[i]表示搶到第i個房子
        #dp[i]=max(dp[i-2]+nums[i],dp[i-1])
        if len(nums)==1:
            return nums[0]
        dp = [0]*(len(nums))
        #初始
        dp[0]=nums[0]
        dp[1]=max(nums[0],nums[1])
        for i in range(2,len(nums)):
            dp[i]=max(dp[i-2]+nums[i],dp[i-1])
        return dp[-1]
        
result = Solution()
nums = [2,7,9,3,1]
ans = result.rob(nums)
print(ans)

## C++:

#include<bits/stdc++.h>
using namespace std;

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

int main(){
    Solution res;
    vector<int> nums={2,7,9,3,1};
    int ans=res.rob(nums);
    cout<<ans<<endl; //12
    return 0;
}