[LC] 198. House Robber
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. Solution 1:
Time: O(N)
Space: O(N)
class Solution:
def rob(self, nums: List[int]) -> int:
if not nums:
return 0
if len(nums) <= 2:
return max(nums) max_lst = [0] * len(nums)
max_lst[0] = nums[0]
max_lst[1] = max(nums[0], nums[1])
for i in range(2, len(nums)):
max_lst[i] = max(max_lst[i - 1], max_lst[i - 2] + nums[i])
return max_lst[-1]
Solution 2:
Time: O(N)
Space: O(1)
class Solution:
def rob(self, nums: List[int]) -> int:
if not nums:
return 0
prev_prev, prev, cur = 0, 0, 0
for num in nums:
prev_prev = prev
prev = cur
cur = max(num + prev_prev, prev)
return cur
[LC] 198. House Robber的更多相关文章
- 198. House Robber,213. House Robber II
198. House Robber Total Accepted: 45873 Total Submissions: 142855 Difficulty: Easy You are a profess ...
- 198. House Robber(动态规划)
198. House Robber You are a professional robber planning to rob houses along a street. Each house ha ...
- leetcode 198. House Robber 、 213. House Robber II 、337. House Robber III 、256. Paint House(lintcode 515) 、265. Paint House II(lintcode 516) 、276. Paint Fence(lintcode 514)
House Robber:不能相邻,求能获得的最大值 House Robber II:不能相邻且第一个和最后一个不能同时取,求能获得的最大值 House Robber III:二叉树下的不能相邻,求能 ...
- [LeetCode] 198. House Robber 打家劫舍
You are a professional robber planning to rob houses along a street. Each house has a certain amount ...
- Leetcode 198 House Robber
You are a professional robber planning to rob houses along a street. Each house has a certain amount ...
- Java for LeetCode 198 House Robber
You are a professional robber planning to rob houses along a street. Each house has a certain amount ...
- (easy)LeetCode 198.House Robber
You are a professional robber planning to rob houses along a street. Each house has a certain amount ...
- 【LeetCode】198 - House Robber
You are a professional robber planning to rob houses along a street. Each house has a certain amount ...
- Java [Leetcode 198]House Robber
题目描述: You are a professional robber planning to rob houses along a street. Each house has a certain ...
随机推荐
- import datetime
import datetimenow = datetime.datetime.now()print('当前时间:',now) 当前时间: 2019-11-21 11:11:58.093122
- 最长特殊序列 II
最长特殊序列 II class Solution { boolean containsSub(String s,String p){ int i,j; for(i=0,j=0;i<p.lengt ...
- MySQL--InnoDB 关键特性
插入缓冲 Insert Buffer 对于非聚集索引的插入或更新操作,不是每一次直接插入到索引页中,而是先判断插入的非聚集索引页是否在缓冲池中,若在,则直接插入:若不在,则先放入到一个 Insert ...
- scala快排
笔记积累,直接看代码吧,sublime上运行的 def quickSort(ls:List[Int]):List[Int]={ if (ls.isEmpty) ls else quickSort(ls ...
- JavaScript学习总结(三)
在学习完了基本的内容之后,我们来学习一下JavaScript中的对象部分以及如何自定义对象的问题. String对象 创建字符串的方式共有两种: 方式1:new String("内容&quo ...
- 大数据攻城狮之进阶技能-Github的使用
引用百度百科中的介绍: github GitHub是一个面向开源及私有软件项目的托管平台,因为只支持git 作为唯一的版本库格式进行托管,故名GitHub. GitHub于2008年4月10日正式上线 ...
- 生产事故(MongoDB数据分布不均解决方案)
可以很明显可以看到我们这个集合的数据严重分布不均匀. 一共有8个分片,面对这个情况我首先想到的是手动拆分数据块,但这不是解决此问题的根本办法. 造成此次生产事故的首要原因就是片键选择上的问题,由于片键 ...
- PAT Basic 1047 编程团体赛(20) [Hash散列]
题目 编程团体赛的规则为:每个参赛队由若⼲队员组成:所有队员独⽴⽐赛:参赛队的成绩为所有队员的成绩和:成绩最⾼的队获胜.现给定所有队员的⽐赛成绩,请你编写程序找出冠军队. 输⼊格式: 输⼊第⼀⾏给出⼀ ...
- ZJNU 1310 - 排队——中高级
蒟蒻做法:追踪1号队员,取他回到原来位置需要的次数 /* Written By StelaYuri */ #include<stdio.h> int main(){ int T,t,n,a ...
- JavaEE--分布式与集群
参考:https://www.zhihu.com/question/20004877 集群是个物理形态,分布式是个工作方式. 只要是一堆机器,就可以叫集群,他们是不是一起协作着干活,这个谁也不知道:一 ...