【leetcode】【198】House Robber
2016-03-15 15:53
501 查看
一、问题描述
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 itwill 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.
二、问题分析
典型的dynamic programing。这种类型的题目,往往需要从头开始判断即初始状态,然后找出状态转移方程。结合我们的题目,当只有一间房子的时候,肯定偷;有两间房子的时候,选择一个钱多的偷;当有三间房子的时候,很明显要么偷第一间和第三间,要么偷第二间;依次类推。对于某间房子,只有两种状态:偷或者不偷。绕过弯来了,就好弄了。三、Java AC代码
public int rob(int[] nums) {int[] res = new int[nums.length];
if (nums==null || nums.length==0) {
return 0;
}
if (nums.length==1) {
return nums[0];
}else {
res[0] = nums[0];
res[1] = Math.max(res[0], nums[1]);
for(int i=2;i<nums.length;i++){
res[i] = Math.max(res[i-2]+nums[i], res[i-1]);
}
return res[nums.length-1];
}
}
相关文章推荐
- java对世界各个时区(TimeZone)的通用转换处理方法(转载)
- java-注解annotation
- java-模拟tomcat服务器
- java-用HttpURLConnection发送Http请求.
- java-WEB中的监听器Lisener
- Android IPC进程间通讯机制
- Android Native 绘图方法
- Android java 与 javascript互访(相互调用)的方法例子
- 介绍一款信息管理系统的开源框架---jeecg
- 聚类算法之kmeans算法java版本
- java实现 PageRank算法
- PropertyChangeListener简单理解
- c++11 + SDL2 + ffmpeg +OpenAL + java = Android播放器
- 插入排序
- 冒泡排序
- 堆排序
- 快速排序
- 二叉查找树