[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： dp

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本文探讨了一种策略，旨在为专业抢劫者提供如何在不触发相邻房屋安全系统警报的情况下，最大化从沿街房屋中窃取的资金。通过动态规划算法，解释了如何使用递归和迭代方法解决此问题，并提供了改进算法以优化时间和空间效率的步骤。

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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.

这是一道比较简单的DP题，递推式是f(n)=max(f(n-1),f(n-2)+num[n]).
一开始写的是纯递归算法，结果TLE了。

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

然后就是应该选择空间换取时间的方法，因为按照上面的方法，f(3)、f(4)、之类的会多次访问，不如用一个数组将它们记录下来。时间复杂度是O(n)

class Solution {
public:
    int ans[10000];
    int rob(vector<int> &num) {
    int n=num.size(),i;
    if(n==0) return 0;
    else if(n==1) return num[0];
    else if(n==2) return max(num[0],num[1]);
    else{
    ans[0] = num[0];
    ans[1] = max(num[0],num[1]);
    for(i=2;i<n;i++){
    ans[i] = max(ans[i-1],num[i]+ans[i-2]);
    }
    return ans[n-1];
    }
    }
};