LeetCode: 198. House Robber

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本文深入解析LeetCode题目198“打家劫舍”，探讨了如何在不触动相邻房屋警报系统的情况下，最大化抢劫金额的策略。通过动态规划方法，详细解释了算法实现过程，提供了清晰的代码示例。

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LeetCode: 198. House Robber

解题思路

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.

解题思路 —— 动态规划

不偷当前房间的最大收益：不偷上一间房间的最大收益和偷上间房间的最大收益的最大值
偷当前房间的最大收益：不偷上一间房间的最大收益加上偷当前房间的收益

AC 代码

class Solution {
public:
    int rob(vector<int>& nums) {
        // first:  当前房间不偷的最大收益
        // second: 偷当前房间的最大收益
        pair<int, int> lastHouse{0, 0}; 
        pair<int, int> curHouse{0, 0}; 

        for(size_t i = 0; i < nums.size(); ++i)
        {
            // 当前房间不偷：上一间房间不偷的最大收益和上间房间偷的最大收益的最大值
            curHouse.first = max(lastHouse.second, lastHouse.first);
            // 偷当前房间：不偷上一间房间的最大收益加上偷当前房间的收益
            curHouse.second = lastHouse.first + nums[i];

            lastHouse = curHouse;
        }

        return max(curHouse.first, curHouse.second);
    }
};