198.House Robber

tags: Medium,Array,DP

198. 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
Marsgoat 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;   
    }
}

Jim 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
Kobe 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

Yen-Chi ChenWed, Dec 14, 2022

Reference

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