198-House Robber_198house robber-优快云博客

本文介绍了一种使用动态规划解决LeetCode上经典问题“打家劫舍”的方法。该问题要求在不触动相邻房屋警报的情况下，计算出能抢劫的最大金额。文章详细解析了算法思路，并给出了C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

类别：dynamic programming
难度：easy

题目描述

https://leetcode.com/problems/house-robber/description/
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.

算法分析

（1）n == 0, return 0
（2）nums[0] = nums[2] = 原值
（3）nums[3] = max(nums[1], nums[0] + nums[2])
（4）nums[i] = max(nums[i] + nums[i - 2], nums[i] + nums[i - 3], nums[i - 1]);

代码实现

//  find maximum sum that do not choose the adjactive value at the same time 
//  leetcode 198
#include <iostream>
#include <vector>
using namespace std;

int rob(vector<int>& nums) {
    int result = 0;
    int n = nums.size();
    if (n == 0) return 0;
    nums[1] = max(nums[0], nums[1]);
    nums[2] = max(nums[1], (nums[0] + nums[2]));
    for (int i = 3; i < n; ++i) {
        nums[i] = max(nums[i] + nums[i - 2], nums[i] + nums[i - 3]);
        nums[i] = max(nums[i], nums[i - 1]);
    }        
    return nums[n - 1];
}

int main() {
    int n;
    int num;
    vector<int> data;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> num;
        data.push_back(num);
    }
    cout << rob(data) << endl;
    return 0;
}