Leetcode 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.

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.

解题思路：这道题的本质相当于在一列数组中取出一个或多个不相邻数，使其和最大。那么对于这类求极值的问题首先考虑动态规划 Dynamic Programming 来解，维护一个一位数组 dp，其中 dp[i] 表示 [0, i] 区间可以抢夺的最大值，对当前i来说，有抢和不抢两种互斥的选择，不抢即为 dp[i-1]（等价于去掉 nums[i] 只抢 [0, i-1] 区间最大值），抢即为 dp[i-2] + nums[i]（等价于去掉 nums[i-1]）。再举一个简单的例子来说明一下吧，比如说 nums为{3, 2, 1, 5}，那么来看 dp 数组应该是什么样的，首先 dp[0]=3 没啥疑问，再看 dp[1] 是多少呢，由于3比2大，所以抢第一个房子的3，当前房子的2不抢，则dp[1]=3，那么再来看 dp[2]，由于不能抢相邻的，所以可以用再前面的一个的 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，这样就可以得到状态转移方程 dp[i] = max(num[i] + dp[i - 2], dp[i - 1]), 且需要初始化 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])，代码如下：

class Solution {
public:
    int rob(vector<int>& nums)
    {
        if(nums.size() <= 1)
            return nums.empty()? 0:nums[0];
        
        int len = nums.size();
        vector<int> dp(len, -1);
        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);

        for(int i = 2; i< len; i++)
        {
            dp[i] = max(dp[i-2] + nums[i], dp[i-1]);
        }
        
        return dp[len-1];
    }
    

};

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来源： https://blog.csdn.net/xuchenhuics/article/details/103943057