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.

给一个数组代表每个房子里的money，一个robber不能rob相邻的房子，问最大能得到多少money

思路：
最大or最小or有多少种方法，dp

对于每个house，robber可以选择rob或者不rob
如果不rob，那么money就保持在前一次的记录dp[i-1]
如果rob，因为不能rob相邻的house，所以只能dp[i-2] + house[i]

这两种方案取较大的作为当前i的money
dp[i] = max(dp[i-1], house[i] + dp[i - 2])

0和1的house比较特殊，因为相邻，所以只能取一个，选最大的那个

    int rob(vector<int>& nums) {
        if (nums.size() == 0) {
            return 0;
        }
        
        if (nums.size() <= 2) {
            return *std::max_element(nums.cbegin(), nums.cend());
        }
     
        vector<int> dp(nums.size(), 0);
        dp[0] = nums[0];
        dp[1] = max(dp[0], nums[1]);

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