leetcode 213. House Robber II

最新推荐文章于 2022-11-01 19:16:12 发布

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最新推荐文章于 2022-11-01 19:16:12 发布

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探讨一个专业窃贼如何在不触动警报的情况下，从排列成环形的房屋中窃取最大金额的问题。窃贼不能连续抢劫相邻房屋，尤其是首尾相连的房屋。通过动态规划算法，使用两个DP数组分别计算包含和不包含第一个房屋的情况，最终确定最大可窃取金额。

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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.

198回顾：
给一个数组代表每个房子里的money，一个robber不能rob相邻的房子，问最大能得到多少money

这题是198题的拓展，198不是环状结构，这道题house是环状结构，也就是第0个和最后一个也算是相邻的，这两个也不能同时rob

思路：
两个dp数组，一个包含第0个house的money，dp1
一个不包含第0个house，dp2

dp1既然包含第0个就不能包含最后一个，所以返回的是
max(dp1[size - 2], dp2[size -1])

    int rob(vector<int>& nums) {
        if (nums.size() == 0) {
            return 0;
        }
        
        if (nums.size() <= 2) {
            return *std::max_element(nums.cbegin(), nums.cend());
        }
        
        vector<int> dp1(nums.size(), 0);
        vector<int> dp2(nums.size(), 0);
        
        dp1[0] = nums[0];
        dp1[1] = max(dp1[0], nums[1]);
        dp2[1] = nums[1];
        
        for (int i = 2; i < nums.size(); i++) {
            dp1[i] = max(dp1[i - 1], nums[i] + dp1[i - 2]);
            dp2[i] = max(dp2[i - 1], nums[i] + dp2[i - 2]);
        }
        
        return max(dp1[nums.size() - 2], dp2[nums.size() - 1]);
    }