Leetcode House Robber

最新推荐文章于 2024-01-18 14:02:27 发布

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最新推荐文章于 2024-01-18 14:02:27 发布

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分类专栏： leetcode 文章标签： leetcode

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本文探讨了一种特殊情况下房屋抢劫的问题求解方法。假设每栋房屋都有一定的现金储备，但相邻房屋的安全系统相互连接，若同一晚两栋相邻房屋被抢劫则会自动报警。文章提供两种算法实现方案，一种使用动态规划数组，另一种采用更简洁的方法，仅用两个变量跟踪前两次的最大值。

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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.

class Solution {
public:
    int dp[10000];
public:
    int rob(vector<int>& nums) {
        if(nums.size()==0) return 0;
        memset(dp,0,sizeof(dp));
        dp[0]=nums[0];
        dp[1]= nums[0]>nums[1]?dp[0]:nums[1];
        for(int i=2;i<nums.size();i++){
            dp[i]=max(dp[i-1],dp[i-2]+nums[i]);
        }
        return dp[nums.size()-1];
    }
};

思考了一下，可以不用数组，遍历到n的时候只要知道dp[n-1]以及dp[n-2]的值就可以了。

class Solution {
public:
    int rob(vector<int>& nums) {
        if(nums.size()==0) return 0;
        int a = 0;
        int b = 0;
        for(int i =0;i<nums.size();i++){
            int tmp=b;
            b=max(b,a+nums[i]);
            a=tmp;
        }
        return b;
    }
};