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.

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.

简单点讲,题目的意思就是,一连串数字,不能去相邻的值,那么怎么样才能取到其中的最大值。DP,表达式为:ret[i] = max(ret[i - 1], ret[i - 2] + nums[i]);

一开始想还有一种可能就是ret[i - 1]可能就是ret[i - 2],但是这里不影响,因为还是ret[i - 2] + num[i]较大,代码如下:

 class Solution {
public:
int rob(vector<int>& nums) {
int sz = nums.size();
vector<int> maxRet = nums;
if(sz == ) return ;
if(sz == ) return nums[];
if(sz == ) return max(nums[], nums[]);
int i;
maxRet[] = nums[];
maxRet[] = max(nums[], nums[]);
for(i = ; i < sz; ++i){
maxRet[i] = max(maxRet[i - ] + nums[i], maxRet[i - ]);
}
return maxRet[i - ];
}
};

java版本的代码和上面一样,如下所示:

 public class Solution {
public int rob(int[] nums) {
int sz = nums.length;
if(sz == 0)
return 0;
if(sz == 1)
return nums[0];
if(sz == 2)
return Math.max(nums[0], nums[1]);
int [] ret = new int [sz];
ret[0] = nums[0];
ret[1] = Math.max(nums[0], nums[1]);
for(int i = 2; i < sz; ++i){
ret[i] = Math.max(ret[i-2] + nums[i], ret[i-1]);
}
return ret[sz-1];
}
}

LeetCode OJ:House Robber(住宅窃贼)的更多相关文章

  1. LeetCode OJ 题解

    博客搬至blog.csgrandeur.com,cnblogs不再更新. 新的题解会更新在新博客:http://blog.csgrandeur.com/2014/01/15/LeetCode-OJ-S ...

  2. 【LeetCode OJ】Interleaving String

    Problem Link: http://oj.leetcode.com/problems/interleaving-string/ Given s1, s2, s3, find whether s3 ...

  3. 【LeetCode OJ】Reverse Words in a String

    Problem link: http://oj.leetcode.com/problems/reverse-words-in-a-string/ Given an input string, reve ...

  4. LeetCode OJ学习

    一直没有系统地学习过算法,不过算法确实是需要系统学习的.大二上学期,在导师的建议下开始学习数据结构,零零散散的一学期,有了链表.栈.队列.树.图等的概念.又看了下那几个经典的算法——贪心算法.分治算法 ...

  5. LeetCode OJ 297. Serialize and Deserialize Binary Tree

    Serialization is the process of converting a data structure or object into a sequence of bits so tha ...

  6. 备份LeetCode OJ自己编写的代码

    常泡LC的朋友知道LC是不提供代码打包下载的,不像一般的OJ,可是我不备份代码就感觉不舒服- 其实我想说的是- 我自己写了抓取个人提交代码的小工具,放在GitCafe上了- 不知道大家有没有兴趣 ht ...

  7. LeetCode OJ 之 Maximal Square (最大的正方形)

    题目: Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and ...

  8. Leetcode 337. House Robber III

    337. House Robber III Total Accepted: 18475 Total Submissions: 47725 Difficulty: Medium The thief ha ...

  9. LeetCode OJ:Integer to Roman(转换整数到罗马字符)

    Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 t ...

  10. LeetCode OJ:Serialize and Deserialize Binary Tree(对树序列化以及解序列化)

    Serialization is the process of converting a data structure or object into a sequence of bits so tha ...

随机推荐

  1. Java中的异常和处理详解(转发:https://www.cnblogs.com/lulipro/p/7504267.html)

    简介 程序运行时,发生的不被期望的事件,它阻止了程序按照程序员的预期正常执行,这就是异常.异常发生时,是任程序自生自灭,立刻退出终止,还是输出错误给用户?或者用C语言风格:用函数返回值作为执行状态?. ...

  2. 剑指offer 面试65题

    题目65题:不用加减乘除做加法. 解法一:Python特性 # -*- coding:utf-8 -*- class Solution: def Add(self, num1, num2): # wr ...

  3. HDU 3199 Hamming Problem

    Hamming Problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tot ...

  4. [JavaScript]常用的页面倒计时

    倒计时是web开发中比较常用的,以下列出常用的几个倒计时方法,仅供参考: 一 :页面倒计时 原理一般都是通过 setTimeout 或 setInterval 函数实现,下面是一个最简单的倒计时 &l ...

  5. Hibernate多对多关联

    多对多关联: 示例:Teacher和Student,一个Teacher可以教很多student,一个Student也可以被很多teacher教   多对多单向关联 Teacher知道自己教了哪些学生, ...

  6. css小技巧1

    资料 1. 文本省略 单行省略: white-space:nowrap; overflow:hidden; text-overflow:ellipsis; 多行文本省略: 只兼容webkit内核,不属 ...

  7. 【HackerRank】Halloween party

    Change language : Alex is attending a Halloween party with his girlfriend Silvia. At the party, Silv ...

  8. ceph安装各种报错

    [ceph_deploy][ERROR ] RuntimeError: Failed to execute command: ceph-disk-activate –mark-init sysvini ...

  9. Ubuntu 使用国内apt源

    编辑/etc/apt/source-list deb http://cn.archive.ubuntu.com/ubuntu/ trusty main restricted universe mult ...

  10. poj 2186 Popular Cows 【强连通分量Tarjan算法 + 树问题】

    题目地址:http://poj.org/problem?id=2186 Popular Cows Time Limit: 2000MS   Memory Limit: 65536K Total Sub ...