[LeetCode] 213. House Robber II

最新推荐文章于 2025-04-14 23:34:52 发布

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最新推荐文章于 2025-04-14 23:34:52 发布

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本文探讨了LeetCode上的环形房屋抢劫者II问题，这是一个经典的动态规划问题。文章详细解释了如何在不触动相邻房屋警报的情况下，计算出从一系列房屋中能够窃取的最大金额。通过对比两个子序列的解决方案，最终得出最大收益。

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

题目

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, 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: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.

Example 2:

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.

思路

题目大意

求一条环路上的所求最大和，求和时，环路两个相邻的元素不能同时计算。

解题思路

基于 198. House Robber 问题的变型。
总体解法还是 House Robber 的dp。最后结果为 nums[0:nums.length-2] 与 nums[1:nums.length-1] 的House Robber 解的较大值。
这是因为 nums[0] 和 nums[nums.length-1] 同时只有一个可以加和。

class Solution {
    public int rob(int[] nums) {
        if(nums.length == 1)    return nums[0];
        return Math.max(eachRob(nums,0,nums.length-2),eachRob(nums,1,nums.length-1));
    }
    public int eachRob(int[] nums,int first,int last){
        int a = 0,b = 0;
        for(int i = first; i<=last;i++){
            int c = Math.max(a+nums[i],b);
            a = b;
            b = c;
        }
        return b;
    }     
}