[LeetCode] HouseRobber

最新推荐文章于 2022-03-05 22:30:26 发布

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最新推荐文章于 2022-03-05 22:30:26 发布

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本文提供了一种高效的策略，用于在不触发相邻房屋安全系统的前提下，最大化从一排房屋中抢劫的收益。通过动态规划方法解决此问题，详细解释了在每个节点的选择过程，并对比了递归方法的不足。最后提供了关键代码实现，包括使用类Solution进行抢劫收益计算。

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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 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.

Credits:

Special thanks to @ifanchu for adding this problem and creating all test cases. Also thanks to @ts for adding additional test cases.

1. DP问题，在每个index i有两个选择：如果抢劫i，则最大收益为(不抢劫i-1)＋val［i］；如果不抢劫i，则最大收益为（抢劫i-1）

class Solution {
public:
    int rob(vector<int>& nums) {
        int rob_true=0;
        int rob_false=0;
        for(auto val: nums)
        {
            int tmp=rob_true;
            //if rob this house, definitely cannot rob the previous house
            rob_true = rob_false + val;
            //if not rob this house, either rob or skip the previous house
            rob_false = max(tmp,rob_false);
        }
        return max(rob_true, rob_false);
    }
};

2. 递归的思路，超时！！！！！

class Solution {
public:
    void helper(vector<int>& nums, int start, int end, int& maxval, int tmpsum)
    {
        if(start>end) return;
        if(start==end){
            tmpsum+=nums[start];
            if(maxval<tmpsum) maxval=tmpsum;
            return;
        }
        //recursively call helper for conditions: (1) if nums[start] is picked;
        //(2) nums[start] is not picked
        helper(nums,start+2,end,maxval,tmpsum+nums[start]);
        helper(nums,start+1,end,maxval,tmpsum); //nums[start] is not picked
    }
    int rob(vector<int>& nums) {
        int maxval=0, tmpsum=0;
        helper(nums,0,nums.size()-1,maxval, tmpsum);
        return maxval;
    }
};