213. House Robber II

最新推荐文章于 2024-05-01 19:56:49 发布

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最新推荐文章于 2024-05-01 19:56:49 发布

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分类专栏： LeetCode Dynamic Programming 文章标签： LeetCode Dynamic Programming House Robber II

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解决一个贼如何在环形排列的房屋中最大化抢劫金额的问题，确保不会因相邻房屋报警而被抓住。通过调整策略，避免首尾房屋同时被抢。

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Note: This is an extension of House Robber.

After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.

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.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

根据上一个题，可以知道怎样解决一排的情况，这道题房子排列是一个圈，那么需要考虑第一个和最后一个的关系。如果第一个偷了，那么最后一个就不能偷，如果第一个没偷，最后一个就可以偷。代码如下：

public class Solution {
    public int rob(int[] nums) {
        if (nums.length == 1) {
            return nums[0];
        }
        return Math.max(robHouse(nums, 0, nums.length - 2), robHouse(nums, 1, nums.length - 1));
    }
    
    public int robHouse(int[] nums, int lo, int hi) {
        int currRob = 0, currNo = 0;
        for (int i = lo; i <= hi; i ++) {
            int preNo = currNo;
            int preRob = currRob;
            currNo = Math.max(preNo, preRob);
            currRob = preNo + nums[i];
        }
        return Math.max(currNo, currRob);
    }
}