198. House Robber

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探讨如何通过动态规划算法解决专业窃贼抢劫问题，在不触动相邻房屋警报的情况下实现利益最大化。文章提供了详细的算法思路及C++实现代码。

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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.
思路：动态规划，这个和币值最大化问题一样的状态转移方程
F(n) = max(c + F(n - 2), F(n - 1)), n > 1

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

从最后抢到多少钱考虑，最后一家抢了达到最大值和最后一家不抢达到最大值。如果最后一家抢了，那么这个最大值=抢了倒数第二家为止达到的最大值+最后一家，最后一家不抢，这个最大值=抢倒数第一家为止达到的最大值
即dp[n] = max(dp[n - 2] + f(n), dp[n - 1);
dp[n - 1]怎么算，也按照上述思路

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