LeetCode 198. House Robber (Easy)

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本文探讨了一道经典的动态规划问题——如何最大化抢劫一组房屋所获得的金钱总额，同时避免触发相邻房屋的安全警报。文章提供了详细的解决方案及C++代码实现。

题目描述：

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.

题目大意：给一个数组，从其中取数，要求和最大，而且相邻两个数字不能同时取。

思路：动态规划，跟期中考的最小和非常像。转移方程：dp[i] = max(nums[i] + dp[i - 2], nums[i] + dp[i - 3])，维护一个最大值记录一下就可以了。

c++代码：

class Solution {
public:
    int rob(vector<int>& nums) {
        int ans = 0;
        if (nums.size() <= 3 && nums.size() > 0)
        {
            if (nums.size() == 1)
                ans = nums[0];
            else if (nums.size() == 2)
                ans = nums[0] > nums[1] ? nums[0] : nums[1];
            else
                ans = nums[0] + nums[2] > nums[1] ? nums[0] + nums[2] : nums[1];
        }
        else if (nums.size() > 3)
        {
            vector<int> dp;
            for (auto i = 0; i < 2; i++)
            {
                dp.push_back(nums[i]);
            }
            dp.push_back(nums[2] + dp[0]);
            ans = dp[2] > dp[1] ? dp[2] : dp[1];
            for (int i = 3; i < nums.size(); i++)
            {
                if (nums[i] + dp[i - 2] > nums[i] + dp[i - 3])
                {
                    dp.push_back(nums[i] + dp[i - 2]);
                }
                else dp.push_back(nums[i] + dp[i - 3]);
                if (ans < dp[i])
                    ans = dp[i];
            }
        }
        return ans;
    }
};