zoukankan      html  css  js  c++  java
  • 198. House Robber,213. House Robber II

    198. House Robber

    Total Accepted: 45873 Total Submissions: 142855 Difficulty: Easy

    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.

     
    class Solution {
    public:
        int rob(vector<int>& nums) {
            int nums_size = nums.size();
            int max_money = 0;
            
            vector<int>dp(nums_size,0);
            
            for(int i=0;i<nums_size;++i){
                int m = 0;
                for(int j=i-2;j>=0;j--){
                    m = max(m,dp[j]);
                }
                dp[i] = m+nums[i];
                max_money = max(max_money,dp[i]);
            }
            return max_money;
        }
        
    };
    
    /**
    dp[i] = max(dp[i-2],dp[i-3]...dp[1])+nums[i];
    [9,8,9,20,8]
    [1,2,3,55,54,2]
    */
     
     

    213. House Robber II

    Total Accepted: 18274 Total Submissions: 63612 Difficulty: Medium

    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.

    /**
    dp[i] = max(dp[i-2],dp[i-3]...dp[1])+nums[i];
    [9,8,9,20,8]
    [1,2,3,55,54,2]
    */
    class Solution {
    public:
        int rob(vector<int>& nums,int start,int end) {
            if(end<=start) return 0;
            int nums_size = end-start;
            int max_money = 0;
            
            vector<int>dp(nums_size,0);
            int k = 0;
            cout<<"start="<<start<<" end="<<endl;
            for(int i=start;i<end;++i){
                int m = 0;
                for(int j=k-2;j>=0;j--){
                    m = max(m,dp[j]);
                }
                dp[k] = m+nums[i];
                cout<<"dp["<<(k)<<"]="<<dp[k]<<endl;
                max_money = max(max_money,dp[k]);
                k++;
            }
            
            return max_money;
        }
        int rob(vector<int>& nums) {
            int nums_size = nums.size();
            return nums_size==1 ? nums[0] : max(rob(nums,0,nums_size-1),rob(nums,1,nums_size));
        }
    };
  • 相关阅读:
    一个爬虫的练习(妹子图)
    安装模块出现的编译问题(解决)
    基于套接字通信的简单练习(FTP)
    Python3 之选课系统
    数据爬取后台(PHP+Python)联合作战
    让pip 使用国内镜像源
    为啥学蛇和python10年后的变化
    模拟登陆百度以及Selenium 的基本用法
    冒泡排序及其效率分析(无聊搞来玩玩)
    LLVM编译器
  • 原文地址:https://www.cnblogs.com/zengzy/p/5060463.html
Copyright © 2011-2022 走看看