动态规划---房子抢劫最大利润

最新推荐文章于 2024-08-05 09:31:45 发布

米缸没米了

最新推荐文章于 2024-08-05 09:31:45 发布

阅读量485

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 文章标签：动态规划

本文链接：https://blog.youkuaiyun.com/qq_31307013/article/details/80576118

LeetCode 专栏收录该内容

41 篇文章

订阅专栏

本文探讨了一个经典的动态规划问题——打家劫舍。该问题要求从一排房屋中选择若干座进行抢劫，每座房屋都有一定的金钱值，但相邻的房屋装有相连的安全系统，若两座相邻的房屋被同时抢劫则会触发报警。文章提供了两种语言（C++ 和 Python）的解决方案，并详细解释了如何通过动态规划求解最大抢劫金额。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1、题目：

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.

Example 1:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

2、解答：为了是f[n]最大化，则必须再选 f[i-2]+nums[i] 和不选 f[i-1] 选择最大值

3、代码：

C++代码

class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size();
        if(n==0)
            return 0;
        if(n==1)
            return nums[0];
        if(n==2)
            return max(nums[0],nums[1]);
        
        vector<int> f(n,0);
        
        f[0] = nums[0];
        f[1] = max(nums[0],nums[1]);
        for(int i=2;i<n;i++){
            f[i] = max(f[i-2]+nums[i],f[i-1]);
        }
        return f[n-1];
        
    }
};

python代码

class Solution:
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        
        dp = [0]*len(nums)
        
        if len(nums) == 1:
            return nums[0]
        
        dp[0] = nums[0]
        dp[1] = max(nums[0],nums[1])
        
        for i in range(2,len(nums)):
            dp[i] = max(dp[i-2]+nums[i],dp[i-1])
            
        return dp[-1]