Leetcode-198-House Robber

最新推荐文章于 2025-08-19 14:43:05 发布

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最新推荐文章于 2025-08-19 14:43:05 发布

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分类专栏：数据结构和算法文章标签：算法动态规划

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本文探讨了一种称为“抢劫者问题”的经典动态规划题目。该问题要求从一系列非负整数中选取不相邻元素以使总和最大化，模拟抢劫者在不能连续抢劫两户的情况下如何计划获取最多财富。

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题目：

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.

int rob(int* nums, int numsSize) {
    int i;
    if(numsSize == 0) return 0;
    if(numsSize>1 && nums[0]>nums[1]) nums[1] = nums[0];
    for(i = 2; i < numsSize; i ++){
        nums[i] = (nums[i-2]+nums[i]>nums[i-1])?(nums[i-2]+nums[i]):nums[i-1];
    }
    return nums[numsSize-1];
}

解析：

题目的意思是给一串非负数，求不相邻的数字最大和（题目中说不能获取相邻两个房子的金钱）

这是一道动态规划题目，我们设定每个房子的金钱为m(k)

到第k个房子时能获得的最大金钱数为p(k)=max(p(k-2)+m(k), p(k-1))

p(0)=0, p(1) = m(1), p(2) = max(p(0)+m(2), p(1))=max(m(2), m(1))