[Leetcode]198. House Robber

最新推荐文章于 2025-09-07 20:32:33 发布

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最新推荐文章于 2025-09-07 20:32:33 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/David-Lin/p/8337805.html

本文探讨了一道经典的动态规划问题——如何最大化抢劫一排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。文章给出了清晰的状态转移方程，并提供了一个C++实现示例。

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：这题的状态转移方程还是比较好找的，然而我转换代码的能力还是太弱，或者说dp题做的还不够多，费了好多时间。。。

方程dp[i] = max{nums[i] + (i -2 < 0 ? 0 : dp[i-2]), dp[i-1]}

 1 class Solution {
 2 public:
 3     int rob(vector<int>& nums) {
 4         if (!nums.size())
 5             return 0;
 6         int dp[nums.size()]; dp[0] = nums[0];
 7         for (int i = 1; i != nums.size(); ++i)
 8             dp[i] = max(nums[i] + (i-2 < 0 ? 0 : dp[i-2]), dp[i-1]);
 9         return dp[nums.size()-1];
10     }
11 };