[leetcode] 198. House Robber

原创

是念 2022-08-11 17:24:55 博主文章分类：C++ ©著作权

©著作权归作者所有：来自51CTO博客作者是念的原创作品，请联系作者获取转载授权，否则将追究法律责任

Description

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.

Example 1:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

分析

题目的意思是：一个小偷取偷东西，但是相邻的房间不能偷，求能够偷到的最大值。

这是一道典型的动态规划的题目。
dp[i]表示到位置i时不相邻数能形成的最大和。

比如说nums为{3, 2, 1, 5}，首先dp[0]=3没啥疑问，再看dp[1]，由于3比2大，所以我们抢第一个房子的3，当前房子的2不抢，所以dp[1]=3，那么再来看dp[2]，由于不能抢相邻的，所以我们可以用在前面的一个的dp值加上当前的房间值，和当前房间的前面一个dp值比较，取较大值当做当前dp值。

所以我们可以得到递推公式dp[i] = max(num[i] + dp[i - 2], dp[i - 1]), 由此看出我们需要初始化dp[0]和dp[1]，其中dp[0]即为num[0]，dp[1]此时应该为max(num[0], num[1])

代码

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