难度: Easy
原题连接
内容描述
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 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.
Example 1:
Input: [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: [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.
思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******
- dp[i][0]代表偷到第i家，并且不偷第i家的最大收益
- dp[i][1]代表偷到第i家，并且偷第i家的最大收益
class Solution:
def rob(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums:
return 0
if len(nums) == 1:
return nums[0]
if len(nums) == 2:
return max(nums)
dp = [[0] * 2 for i in range(len(nums))]
dp[0][1] = nums[0]
for i in range(1, len(nums)):
dp[i][0] = max(dp[i-1][0], dp[i-1][1])
dp[i][1] = dp[i-1][0] + nums[i]
return max(dp[-1])思路 2 - 时间复杂度: O(N)- 空间复杂度: O(N)******
可以优化一下
dp[i]代表只偷到第i家的maximum amount of money
状态转移方程:
dp[i] = max(dp[i-1], dp[i-2] + nums[i])
beats 100%
class Solution:
def rob(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
n = len(nums)
if n == 0:
return 0
elif n == 1:
return nums[0]
elif n == 2:
return max(nums[0], nums[1])
else:
dp = [0 for i in range(n)]
dp[0] = nums[0]
dp[1] = max(nums[0],nums[1])
for i in range(2, n):
dp[i] = max(dp[i-1], dp[i-2] + nums[i])
return dp[-1]思路 3 - 时间复杂度: O(N)- 空间复杂度: O(1)******
迭代
class Solution(object):
def rob(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
last, now = 0, 0
for num in nums:
last, now = now, max(last+num, now)
return now