198. House Robber

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 cal(vector<int>& v, vector<int>& p, int x){
        if(x>=0){
            if(p[x] != -1) return p[x];
            p[x] = max(cal(v, p, x-2) + v[x],  cal(v, p, x-1));
            return p[x];
        }else{
            return 0;
        }
    }
    
    int rob(vector<int>& nums) {
        vector<int> p = vector<int>();
        p.resize(nums.size());
        for(int i=0;i<nums.size();i++) p[i] = -1;
        return cal(nums, p, nums.size()-1);
    }
};
最后編輯于
?著作權(quán)歸作者所有,轉(zhuǎn)載或內(nèi)容合作請(qǐng)聯(lián)系作者
【社區(qū)內(nèi)容提示】社區(qū)部分內(nèi)容疑似由AI輔助生成,瀏覽時(shí)請(qǐng)結(jié)合常識(shí)與多方信息審慎甄別。
平臺(tái)聲明:文章內(nèi)容(如有圖片或視頻亦包括在內(nèi))由作者上傳并發(fā)布,文章內(nèi)容僅代表作者本人觀點(diǎn),簡(jiǎn)書系信息發(fā)布平臺(tái),僅提供信息存儲(chǔ)服務(wù)。

相關(guān)閱讀更多精彩內(nèi)容

友情鏈接更多精彩內(nèi)容