Dynamic Programming - 213. House Robber II
213. House Robber II You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, 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: [2,3,2] Output: 3 Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.
Example 2:
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.
思路:
这一题和198题多了一个条件就是房子是排列成一个圆形,也就是说,在前一题的基础上,只能从
0
偷到len(nums)-2
或者从1
偷到len(num)-1
。也是用动态规划求解。
代码:
func rob(nums []int) int {
if nums == nil || len(nums) == 0{return 0}
leng := len(nums)
if leng == 1 {return nums[0]}
return max(helper(nums, 0, leng-2), helper(nums, 1, leng-1))
}
func helper(input []int, start int, end int) int {
nums := input[start:end+1]
if nums == nil || len(nums) == 0{ return 0 }
leng := len(nums)
if leng == 1 {return nums[0]}
dp := []int{nums[0], max(nums[0], nums[1])}
index := 0
for i := 2; i < leng; i++ {
index = i % 2 //index^1: 0->1, 1->0.
dp[index] = max(dp[index] + nums[i], dp[index^1])
}
return max(dp[0], dp[1])
}
func max(i, j int) int {
if i > j {
return i
}
return j
}
- JavaScript 教程
- JavaScript 编辑工具
- JavaScript 与HTML
- JavaScript 与Java
- JavaScript 数据结构
- JavaScript 基本数据类型
- JavaScript 特殊数据类型
- JavaScript 运算符
- JavaScript typeof 运算符
- JavaScript 表达式
- JavaScript 类型转换
- JavaScript 基本语法
- JavaScript 注释
- Javascript 基本处理流程
- Javascript 选择结构
- Javascript if 语句
- Javascript if 语句的嵌套
- Javascript switch 语句
- Javascript 循环结构
- Javascript 循环结构实例
- Javascript 跳转语句
- Javascript 控制语句总结
- Javascript 函数介绍
- Javascript 函数的定义
- Javascript 函数调用
- Javascript 几种特殊的函数
- JavaScript 内置函数简介
- Javascript eval() 函数
- Javascript isFinite() 函数
- Javascript isNaN() 函数
- parseInt() 与 parseFloat()
- escape() 与 unescape()
- Javascript 字符串介绍
- Javascript length属性
- javascript 字符串函数
- Javascript 日期对象简介
- Javascript 日期对象用途
- Date 对象属性和方法
- Javascript 数组是什么
- Javascript 创建数组
- Javascript 数组赋值与取值
- Javascript 数组属性和方法
- 下载歌曲的时候嫌麻烦?打造专属你的音乐下载器
- Tidyverse补充
- 抖音关键词热度搜索小程序(附源码)
- python自定义函数基础
- Python-科学计算-pandas-13-列名/删除列/替换nan
- python小程序,45行代码实现可切换版代码雨(附源码)
- R海拾遗-stringr
- stringr2
- Kubernetes 无状态应用的一般特征
- 一段简单的代码,能让所有GIF图实现时光倒流
- 你喜欢的女主播颜值多少分,今天带你测试虎牙直播女主播的颜值
- (译)kubectl 的奇技淫巧
- Python实现主播人气排行榜,带你发现人气王
- (译)Kubernetes:移除 CPU 限制,服务运行更快
- python_不误正业之贪吃蛇