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. Hibernate监听器

    Hibernate的事件监听机制 Hibernate中的事件监听机制可以对Session对象的动作进行监听,一旦发生了特殊的事件,Hibernate就会执行监听器中的事件处理方法 在某些功能的设计中, ...

  2. codeforces 624A Save Luke(水题)

    A. Save Luke time limit per test 1 second memory limit per test 256 megabytes input standard input o ...

  3. 使用friso中文分词注意

    friso是使用c语言开发的一款中文分词器,使用流行的mmseg算法实现.完全基于模块化设计和实现,可以很方便的植入到其他程序中,例如:MySQL,PHP等.目前只支持UTF-8 编码. 源码无需修改 ...

  4. Codeforces Round #271 (Div. 2) D. Flowers (递推)

    题目链接:http://codeforces.com/problemset/problem/474/D 用RW组成字符串,要求w的个数要k个连续出现,R任意,问字符串长度为[a, b]时,字符串的种类 ...

  5. C#一些知识点:委托和事件的区别

    在C#中,委托和事件是比较容易混淆的两个知识点,本篇博客就记录一下委托和事件之间的区别. 定义上的区别 委托:委托实际上是一个类,用来表示一个函数,可以理解为C++中的函数指针. 事件:事件是一个修饰 ...

  6. WordPress主题制作教程[壹] - 了解WP&结构&索引

    最近开始筹备WordPress主题开发了.首先我们在此章节中进行了解什么是WP,以及WP的结构.通过这个文章索引到以后所写的WP系列教程. (抱歉,大家不要急,持续更新中....) 1.首先,我们来认 ...

  7. AOE关键路径

    这个算法来求关键路径,其实就是利用拓扑排序,首先求出,每个节点最晚开始时间,再倒退求每个最早开始的时间. 从而算出活动最早开始的时间和最晚开始的时间,如果这两个时间相等,则为关键路径. 时间复杂度为O ...

  8. 【转】安装Intel HAXM为Android 模拟器加速,30秒内启动完成

    http://www.cnblogs.com/Li-Cheng/p/4351966.html http://www.cnblogs.com/csulennon/p/4178404.html https ...

  9. 【转】Dijkstra算法(单源最短路径)

    原文:http://www.cnblogs.com/dolphin0520/archive/2011/08/26/2155202.html 单源最短路径问题,即在图中求出给定顶点到其它任一顶点的最短路 ...

  10. BZOJ 2748: [HAOI2012]音量调节 dp

    2748: [HAOI2012]音量调节 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.lydsy.com/JudgeOnline/p ...