198.House Robber
===

###### tags: `Medium`,`Array`,`DP`

[198. House Robber](https://leetcode.com/problems/house-robber/)

### 題目描述

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***.

### 範例

**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

### 解答

#### Javascript

解法一
Time: $O(n)$
Space: $O(n)$

function rob(nums) {
  if (nums.length === 0) return 0;
  if (nums.length === 1) return nums[0];
  if (nums.length === 2) return Math.max(nums[0], nums[1]);

  const dp = [];
  dp[0] = nums[0];
  dp[1] = Math.max(nums[0], nums[1]);
  for (let i = 2; i < nums.length; i++) {
    dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
  }

  return dp[nums.length - 1];
}

解法二
Time: $O(n)$
Space: $O(1)$

function rob(nums) {
  let max = 0;
  let prevMax = 0;
  for (let i = 0; i < nums.length; i++) {
    const temp = max;
    max = Math.max(prevMax + nums[i], max);
    prevMax = temp;
  }

  return max;
}

> 解法一是好久以前寫的，真的是圖樣圖森破
> [name=Marsgoat][time= Dec 14, 2022]

#### C#

public class Solution {
    public int Rob(int[] nums) {
        if (nums.Length == 1) return nums[0];

        int amount1 = nums[0];
        int amount2 = Math.Max(nums[0], nums[1]);
        
        for(int i = 2; i < nums.Length; i++){
            int amount = Math.Max(amount2, amount1 + nums[i]);
            amount1 = amount2;
            amount2 = amount;
        }
        
        return amount2;   
    }
}

> [name=Jim][time= Dec 14, 2022]

#### Python

class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums) == 1: return nums[0]
        if len(nums) == 2: return max(nums[0], nums[1])

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

        for i in range(2, len(nums)):
            dp.append(max(dp[i-2] + nums[i], dp[i-1]))

        return dp[-1]

> General Solution

class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums) == 1: return nums[0]
        if len(nums) == 2: return max(nums[0], nums[1])

        first, second = nums[0], max(nums[0], nums[1])

        for i in range(2, len(nums)):
            curr = max(first + nums[i], second)
            first = second
            second = curr

        return second

> Constant Space Solution
> [name=Kobe][time= Dec 14, 2022]

class Solution:
    def rob(self, nums: List[int]) -> int:
        cur, pre = 0, 0
        for n in nums:
            cur, pre = max(cur, pre + n), cur
        return cur

> [name=Yen-Chi Chen][time=Wed, Dec 14, 2022]
### Reference

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