【Leetcode 198】关关的刷题日记69 – Leetcode 198 House Robber

关关的刷题日记69 – Leetcode 198 House Robber

题目

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.

题目的意思是小偷沿着一排街道抢劫房子，每个房子都有一定数额的钱，但是不可以抢相邻的房子，问可以抢到的最大的钱数。

思路

动态规划的题目：设re[i]是抢劫前i个房子获得的最大收益，那么可以分为第i个房子不被抢和第i个房子被抢两种情况：第i个房子不被抢，re[i]=re[i-1]；第i个房子被抢，re[i]=re[i-2]+nums[i]；状态转移方程re[i]=max(re[i-1], re[i-2]+nums[i])。

class Solution {public:
    int rob(vector<int>& nums) {
        int n=nums.size();
        if(n==0)
            return 0;
        if(n==1)
            return nums[0];
        if(n==2)
            return max(nums[0], nums[1]);
        int temp1=nums[0], temp2=max(nums[0],nums[1]), re;
        for(int i=2; i<n; i++)
        {
            re=max(temp1+nums[i],temp2);
            temp1=temp2;
            temp2=re;
        }
        return re;
    }};

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