198. House Robber && 213. House Robber II

专业抢劫者的算法策略

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本文探讨了一种特殊的抢劫场景——沿街抢劫每户有一定金额现金的房子，但不能连续抢劫相邻的两户。通过动态规划的方法，设计了一个高效的算法来确定最大可抢金额，并给出了解决方案的具体实现。

问题描述
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.

Credits:
Special thanks to @ifanchu for adding this problem and creating all test cases. Also thanks to @ts for adding additional test cases.

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解决思路
设置dp[n+1][2]的状态数组，dp[i][0]表示不拿i的时候，dp[i][1]表示拿i的时候，则有以下状态转移方程：
```
dp[i][0] = max(dp[i-1][0],dp[i-1][1])
dp[i][1] = dp[i-1][1]+nums[i-1]
```
很显然，这里只有i和i-1的关系，可以做空间压缩，只需要两个常用变量就可以了。
213题中，数组变成了环，只需要分开0…n-1 1….n这两种情况处理一下就可以了。
代码
代码是213的

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