leetcode之House Robber

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于 2022-04-10 15:21:15 发布

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本文链接：https://blog.youkuaiyun.com/WIMPs_L/article/details/124071407

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动态规划编程问题抢劫问题状态转移最大和

关键词由优快云通过智能技术生成

这篇博客探讨了一个关于使用动态规划解决编程问题的例子，即如何在不触发警报的情况下最大化夜间抢劫房屋的总金额。文章通过分析示例输入和输出，强调了在处理这类问题时不能简单地比较相邻元素的奇偶组合。解决方案中，博主给出了C++实现的代码，该代码通过动态规划状态转移方程求解最大值。注意，在某些情况下，如[2,1,1,2]，最大和并非来自奇偶项交替选取，而是选择最大的连续子序列。

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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 systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: nums = [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: nums = [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.

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 400

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