leetcode:198 House Robber－每日编程第二十二题

最新推荐文章于 2020-04-20 17:34:58 发布

寒风中的易水

最新推荐文章于 2020-04-20 17:34:58 发布

阅读量642

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/fjzpdkf/article/details/50281517

leetcode 专栏收录该内容

41 篇文章

订阅专栏

这篇文章介绍了一个专业的抢劫者如何在一条街道上进行抢劫，同时避免触发相邻房屋的安保系统。通过分析每个房屋内藏匿的财富，文章提供了一种策略来最大化抢夺的金额，而不会被警方发现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

House Robber

Total Accepted: 44999 Total Submissions: 140629 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.

思路：arr[i]表示以第i个被偷为前提，前i＋1个屋子最多可以偷多少钱。

arr[i]=max(arr[i-2],arr[i-3])+num[i];

class Solution {
public:
    int rob(vector<int>& nums) {
        int size = nums.size();
        if(size==0){
            return 0;
        }else if(size==1){
            return nums[0];
        }else if(size==2){
            return (nums[0]>nums[1]?nums[0]:nums[1]);
        }else if(size==3){
            return ((nums[1]>nums[0]+nums[2])?nums[1]:(nums[0]+nums[2]));
        }
        int* arr = new int[size];
        
        arr[0]=nums[0];
        arr[1]=nums[1];
        arr[2]=nums[0]+nums[2];
        
        for(int i=3;i<size;i++){
            arr[i]=nums[i]+(arr[i-2]>arr[i-3]?arr[i-2]:arr[i-3]);
        }
        return (arr[size-1]>arr[size-2]?arr[size-1]:arr[size-2]);
        
    }
};