House Robber
2016-03-09 16:25
155 查看
1 House Robber
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.
public class Solution { public int rob(int[] nums) { int n = nums.length; // 没有房子的时候收入是0 if (n == 0) return 0; // 只有一个房子的时候选择唯一 else if (n == 1) { return nums[0]; } else { int[] maxRob = new int ; maxRob[0] = nums[0]; // 当有两个房子的时候,只能二选一,当然选钱多的那一个 maxRob[1] = Math.max(nums[0], nums[1]); // 房子从0开始编号 // 当考虑是否抢劫第i号房子的时候,有两种情况: // (1)抢i,则i-1不能抢,最大收益是 nums[i] + maxRob[i-2] // (2)不抢i,则最大收益是maxRob[i-1] for (int i = 2; i < n; i++) { maxRob[i] = Math.max(nums[i] + maxRob[i - 2], maxRob[i - 1]); } return maxRob[n - 1]; } } }
2 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.
public class Solution { public int rob(int[] nums) { int n = nums.length; if (n == 0) return 0; else if (n == 1) { return nums[0]; } else { //如果抢最后一个房子,则第一个房子不能抢; //最大收益为,rob1中,抢1到n-1号房子的最大收益 int[] nums1 = Arrays.copyOfRange(nums, 1, n); int maxNoFirst = rob1(nums1); //如果不抢最后一个房子,则第一个房子可以抢; //最大收益为,rob1中,抢0到n-2号房子的最大收益 int[] nums2 = Arrays.copyOfRange(nums, 0, n-1); int maxNoLast = rob1(nums2); return Math.max(maxNoFirst, maxNoLast); } } public int rob1(int[] nums) { int n = nums.length; // 没有房子的时候收入是0 if (n == 0) return 0; // 只有一个房子的时候选择唯一 else if (n == 1) { return nums[0]; } else { int[] maxRob = new int ; maxRob[0] = nums[0]; // 当有两个房子的时候,只能二选一,当然选钱多的那一个 maxRob[1] = Math.max(nums[0], nums[1]); // 房子从0开始编号 // 当考虑是否抢劫第i号房子的时候,有两种情况: // (1)抢i,则i-1不能抢,最大收益是 nums[i] + maxRob[i-2] // (2)不抢i,则最大收益是maxRob[i-1] for (int i = 2; i < n; i++) { maxRob[i] = Math.max(nums[i] + maxRob[i - 2], maxRob[i - 1]); } return maxRob[n - 1]; } } }
相关文章推荐
- ArcGis 给数据库追加Domain
- 143.View the Exhibit and examine the structure of the PROMOTIONS table.
- win10桌面背景为什么突然变黑了 win10桌面背景不显示解决方法
- spark on yarn 配置history server
- Message-oriented middleware
- EditText的功能与用法
- lvs持久性工作原理和配置
- log4j.xml的写法
- 状态模式
- VBA教程初级(五):复杂数组
- 约瑟夫环问题
- 分布式Web服务器架构
- python脚本实现文件夹增量复制
- json常用jar包
- Java虚拟机的类加载机制
- Git
- [网络流24题][CODEVS1237]餐巾计划问题(费用流)
- c++继承与组合
- 论MOBA类游戏五号位的重要性
- 10g RAC的dataguad异常一则