LeetCode 213. House Robber II

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.

【题目分析】

在House Robber的基础上，这个题目增加的一个限制就是所有的房屋构成了一个环，在这种情况下保证任不偷盗任意两间相邻的房子。

【思路】

看了别人的好多解法，其实说的都不明不白，我把我的想法分享给大家，肯定让你豁然开朗。

首先：如果房屋不构成一个圈，我们的解法有如下几种可能

1. 头部和尾部的房屋都被抢劫

2. 头部房屋被抢劫，尾部房屋没有被抢劫

3. 头部房屋没有被抢劫，尾部的房屋被抢劫

4. 头部和尾部的房屋都没有被抢劫

如果房屋形成一个环，我们的最优解就只能选择后面的三种情况，第二种情况我们要保证最后一个房屋不能被抢劫，第三种情况要保证第一个房屋不能被抢劫，第四种情况已经包含在前两种情况中了。其实就是在环的连接处保证其中一个不被偷的情况下，求出从剩下的房屋中最多能盗取的金钱数目。

因此在House Robber的基础上我们只要保证结果只能为上面第二和第三种情况即可。代码如下：

 public class Solution {
     public int rob(int[] nums) {
         int len = nums.length;
         if(len == 0) return 0;
         if(len == 1) return nums[0];

         int post2 = nums[len-1];
         int post1 = Math.max(nums[len-1], nums[len-2]);
         int lable1 = 0;
         //保证为第一个房屋不被偷
         for(int i = len-3; i >= 1; i--) {
             int temp = post1;
             post1 = Math.max(post1, nums[i] + post2);
             post2 = temp;
         }
         int result1 = post1;

         post2 = 0;
         post1 = nums[len-2];
         //保证最后一个房屋不被偷
         for(int i = len-3; i >= 0; i--) {
             int temp = post1;
             post1 = Math.max(post1, nums[i] + post2);
             post2 = temp;
         }
         int result2 = post1;

         return Math.max(result1, result2);
     }
 }