House Robber II

Note: This is an extension of House Robber.

After robbing those houses on that street,
the thief has found himself a new place for his thievery
so that he will not get too much attention.
This time, all houses at this place are arranged in a circle.
That means the first house is the neighbor of the last one.
Meanwhile, the security system for these houses remain the same as
for those in the previous street.

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.

 /*************************************************************************
> File Name: LeetCode213.c
> Author: Juntaran
> Mail: JuntaranMail@gmail.com
> Created Time: Wed 11 May 2016 17:11:02 PM CST
************************************************************************/ /************************************************************************* House Robber II Note: This is an extension of House Robber. After robbing those houses on that street,
the thief has found himself a new place for his thievery
so that he will not get too much attention.
This time, all houses at this place are arranged in a circle.
That means the first house is the neighbor of the last one.
Meanwhile, the security system for these houses remain the same as
for those in the previous street. 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. ************************************************************************/ #include <stdio.h> /*
跟198的区别在于现在是一个环形,首尾不能同时get
所以分成两种情况:
1:从头取到尾-1
2:从头+1取到尾
rob过程相同,然后比较两种情况大小
*/
int rob( int* nums, int numsSize )
{
if( numsSize == )
{
return ;
}
if( numsSize == )
{
return nums[];
} int max = ;
int prev1 = ;
int prev2 = ; int i, temp; temp = ;
prev1 = ;
prev2 = ;
for( i=; i<=numsSize-; i++ )
{
temp = prev1;
prev1 = (prev2+nums[i])>prev1 ? (prev2+nums[i]) : prev1;
prev2 = temp;
}
max = prev1; temp = ;
prev1 = ;
prev2 = ;
for( i=; i<=numsSize-; i++ )
{
temp = prev1;
prev1 = (prev2+nums[i])>prev1 ? (prev2+nums[i]) : prev1;
prev2 = temp;
}
max = max>prev1 ? max : prev1; return max;
} int main()
{
int nums[] = { ,,,,,, };
int numsSize = ; int ret = rob( nums, numsSize );
printf("%d\n", ret);
return ;
}

LeetCode 213的更多相关文章

  1. [LeetCode] 213. House Robber II 打家劫舍之二

    You are a professional robber planning to rob houses along a street. Each house has a certain amount ...

  2. [LeetCode] 213. House Robber II 打家劫舍 II

    Note: This is an extension of House Robber. After robbing those houses on that street, the thief has ...

  3. Java实现 LeetCode 213 打家劫舍 II(二)

    213. 打家劫舍 II 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.这个地方所有的房屋都围成一圈,这意味着第一个房屋和最后一个房屋是紧挨着的.同时,相邻的房屋装有相互连通的防盗 ...

  4. Java for LeetCode 213 House Robber II

    Note: This is an extension of House Robber. After robbing those houses on that street, the thief has ...

  5. 【LeetCode 213】House Robber II

    This is an extension of House Robber. After robbing those houses on that street, the thief has found ...

  6. LeetCode 213. House Robber II

    Note: This is an extension of House Robber. After robbing those houses on that street, the thief has ...

  7. leetcode 213. 打家劫舍 II JAVA

    题目: 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.这个地方所有的房屋都围成一圈,这意味着第一个房屋和最后一个房屋是紧挨着的.同时,相邻的房屋装有相互连通的防盗系统,如果两间相邻 ...

  8. Leetcode 213.大家劫舍II

    打家劫舍II 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.这个地方所有的房屋都围成一圈,这意味着第一个房屋和最后一个房屋是紧挨着的.同时,相邻的房屋装有相互连通的防盗系统,如果两 ...

  9. [leetcode] #213 House Robber II Medium (medium)

    原题链接 比子母题House Robber多了一个条件:偷了0以后,第n-1间房子不能偷. 转换思路为求偷盗[0,n-1)之间,以及[1,n)之间的最大值. 用两个DP,分别保存偷不偷第0间房的情况. ...

随机推荐

  1. 【转】Maven实战(五)---两个war包的调用

    原博文出自于: http://blog.csdn.net/liutengteng130/article/details/42879803    感谢! 开篇前提   1.为什么要用两个war包的调用? ...

  2. 我的github

    我的github:先来贴个图~   这是我的github,新建了第一个repository,默认路径是aokoqingiz/code. 然后是里面的文件~ 里面有一个readme.txt,是我对这个r ...

  3. ISP与IAP的区别

    转: ISP(In-System Programming)在系统可编程,指电路板上的空白器件可以编程写入最终用户代码, 而不需要从电路板上取下器件,已经编程的器件也可以用ISP方式擦除或再编程.IAP ...

  4. poj 1847 Tram

    http://poj.org/problem?id=1847 这道题题意不太容易理解,n个车站,起点a,终点b:问从起点到终点需要转换开关的最少次数 开始的那个点不需要转换开关 数据: 3 2 1// ...

  5. Spring Data JPA教程,第一部分: Configuration(翻译)

    Spring Data JPA项目旨在简化基于仓库的JPA的创建并减少与数据库交互的所需的代码量.本人在自己的工作和个人爱好项目中已经使用一段时间,它却是是事情如此简单和清洗,现在是时候与你分享我的知 ...

  6. Django 使用原生SQL

    def dictfetchall(cursor): "将游标返回的结果保存到一个字典对象中" desc = cursor.description return [ dict(zip ...

  7. EasyUI datagrid添加右键菜单项

    js代码 //动态加载数据表格 function InitData() { $('#grid').datagrid({ url: '/Home/Query?r=' + Math.random(), / ...

  8. 《JavaScript高级程序设计》 读书笔记(三)

    操作符 递增和递减操作符 var num1 = 2; var num2 = 20; var num3 = --num1 + num2; // 等于 21 var num4 = num1 + num2; ...

  9. 让div变得大方美观 bootstrap

    <div class="panel panel-default "> <div class="panel-heading"> <h ...

  10. 【M15】了解异常处理(exception handling)的成本

    1.为了在运行期处理异常,程序必须做大量额外的工作.比如,即使抛出异常,也必须保证离开作用域的栈上对象执行析构方法.因此,必须记录try语句的进入点和离开点,记录catch语句能够处理的异常等.这就意 ...