213 House Robber II

最新推荐文章于 2021-09-27 21:36:25 发布

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探讨了一个专业窃贼如何在不触动警报的情况下，从沿街排列成环形的房屋中盗窃最大金额的问题。通过分析相邻房屋的安全系统连接，提出了解决方案，并提供了详细的算法实现。

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1 题目

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, 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: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.
Example 2:

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.

2 尝试解

class Solution {
public:
    int rob(vector<int>& nums) {
        if(nums.size()==0) return 0;
        if(nums.size()==1) return nums[0];
        if(nums.size()==2) return max(nums[0],nums[1]);
        vector<int> cycle1(nums.begin(),nums.end()-1);
        vector<int> cycle2(nums.begin()+1,nums.end());
        return max(uncyclerob(cycle1),uncyclerob(cycle2));
    }
    int uncyclerob(vector<int> nums){
        if(nums.size()==0) return 0;
        if(nums.size()==1) return nums[0];
        if(nums.size()==2) return max(nums[0],nums[1]);
        if(nums.size()==3) return max(nums[1],nums[0]+nums[2]);
        vector<int> saver(nums.size());
        saver[0] = nums[0];
        saver[1] = nums[1];
        saver[2] = nums[2]+nums[0];
        for(int i = 3; i < nums.size();i++){
            saver[i] = max(saver[i-2],saver[i-3])+nums[i];
        }
        return max(saver.back(),saver[saver.size()-2]);
    }
};

3 标准解

class Solution {
public:

    int robOriginal(vector<int>& nums) {
        int a = 0, b = 0, res = 0;
        
        for(int i = 0; i < nums.size(); ++i){
            res = max(b + nums[i], a);
            b = a;
            a = res;
        }
        
        return res;
    }

    int rob(vector<int>& nums) {
        if(nums.empty()) return 0;
        if(nums.size() == 1) return nums[0];
        
        vector<int> numsA(nums.begin() + 1, nums.end());
        vector<int> numsB(nums.begin(), nums.end()-1);
        
        return max(robOriginal(numsA), robOriginal(numsB));
    }
};