LeetCode 198. House Robber

## 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 systems connected and **it will automatically contact the police if two adjacent houses were broken into on the same night.**

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: nums = [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: nums = [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.

class Solution {
public:
    int rob(vector<int>& nums) {
        //nums[i] 表示到第i件当前能偷到的最大价值量
        
        int n=nums.size();
        int f[n+5];
        if(n==1)return nums[0];
        else if(n==2)return max(nums[0],nums[1]);
        else if(n==3)return max(nums[0]+nums[2],nums[1]);
        f[1]=nums[0];
        f[2]=max(nums[0],nums[1]);
        f[3]=max(nums[0]+nums[2],nums[1]);
        for(int i=4;i<=n;i++){
            f[i]=max(f[i-1],f[i-2]+nums[i-1]);
        }
        
        return f[n];
    }
};

标签：nums,house,House,rob,int,amount,money,LeetCode,Robber
来源： https://www.cnblogs.com/zsj1181310894/p/14691557.html