LeetCode 198 House Robber

LeetCode

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

算法思路：

N/A

注意事项：

1. 循环不是模板中的1开始，而是从2开始，因为i-2>=0

Python代码：

def rob(self, nums: List[int]) -> int:
	dp = [0] * (len(nums) + 1)
	dp[1] = nums[0]
	for i in range(2, len(dp)):
		dp[i] = max(dp[i-1], dp[i-2] + nums[i - 1])
	return dp[-1]

算法分析：

时间复杂度为O(n)，空间复杂度O(1)