鱼C论坛

 找回密码
 立即注册
查看: 1914|回复: 0

[学习笔记] Leetcode 198. House Robber

[复制链接]
发表于 2020-7-30 12:15:13 | 显示全部楼层 |阅读模式

马上注册,结交更多好友,享用更多功能^_^

您需要 登录 才可以下载或查看,没有账号?立即注册

x
本帖最后由 Seawolf 于 2020-7-30 12:29 编辑
  1. 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.

  2. 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.



  3. Example 1:

  4. Input: nums = [1,2,3,1]
  5. Output: 4
  6. Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
  7.              Total amount you can rob = 1 + 3 = 4.
  8. Example 2:

  9. Input: nums = [2,7,9,3,1]
  10. Output: 12
  11. Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
  12.              Total amount you can rob = 2 + 9 + 1 = 12.


  13. Constraints:

  14. 0 <= nums.length <= 100
  15. 0 <= nums[i] <= 400
复制代码

  1. 你是一个专业的小偷,计划偷窃沿街的房屋。每间房内都藏有一定的现金,影响你偷窃的唯一制约因素就是相邻的房屋装有相互连通的防盗系统,如果两间相邻的房屋在同一晚上被小偷闯入,系统会自动报警。

  2. 给定一个代表每个房屋存放金额的非负整数数组,计算你 不触动警报装置的情况下 ,一夜之内能够偷窃到的最高金额。


  3. 示例 1:

  4. 输入:[1,2,3,1]
  5. 输出:4
  6. 解释:偷窃 1 号房屋 (金额 = 1) ,然后偷窃 3 号房屋 (金额 = 3)。
  7. 偷窃到的最高金额 = 1 + 3 = 4 。
  8. 示例 2:

  9. 输入:[2,7,9,3,1]
  10. 输出:12
  11. 解释:偷窃 1 号房屋 (金额 = 2), 偷窃 3 号房屋 (金额 = 9),接着偷窃 5 号房屋 (金额 = 1)。
  12. 偷窃到的最高金额 = 2 + 9 + 1 = 12 。


  13. 提示:

  14. 0 <= nums.length <= 100
  15. 0 <= nums[i] <= 400
复制代码


Solution 1:
  1. class Solution:
  2.     def rob(self, nums: List[int]) -> int:
  3.         steal = [0 for _ in nums]
  4.         cooldown = [0 for _ in nums]
  5.         res = 0
  6.         for i in range(len(nums)):
  7.             steal[i] = cooldown[i - 1] + nums[i]
  8.             cooldown[i] = max(cooldown[i - 1], steal[i - 1])
  9.             res = max(steal[i], cooldown[i])
  10.         return res
复制代码


Solution 2:
  1. class Solution:
  2.     def rob(self, nums: List[int]) -> int:
  3.         dp = [0 for _ in nums]
  4.         res = 0
  5.         if len(nums) == 1:
  6.             return nums[0]
  7.         if len(nums) == 2:
  8.             return max(nums)
  9.         for i in range(len(nums)):
  10.             dp[i] = max(dp[i - 1], nums[i] + dp[i - 2])
  11.             res = max(res, dp[i])
  12.         return res
复制代码

本帖被以下淘专辑推荐:

想知道小甲鱼最近在做啥?请访问 -> ilovefishc.com
回复

使用道具 举报

您需要登录后才可以回帖 登录 | 立即注册

本版积分规则

小黑屋|手机版|Archiver|鱼C工作室 ( 粤ICP备18085999号-1 | 粤公网安备 44051102000585号)

GMT+8, 2024-4-19 15:31

Powered by Discuz! X3.4

© 2001-2023 Discuz! Team.

快速回复 返回顶部 返回列表