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.

这道题和之前那道题有些类似，唯一的区别在于这次房屋由直线型变为环形，也就是说第一个和最后一个房屋不能同时抢劫，那我们这里变通一下，如果我们把第一家和最后一家分别去掉，各算一遍能抢的最大值，然后比较两个值取其中较大的一个即为所求。所以这道题可以利用上一题的结果，取抢劫[0,n-2] 和 [1,n-1]之间结果的最大值，代码如下：

 1 class Solution {
 2 public:
 3     int rob(vector<int>& nums) {
 4         int n = nums.size();
 5         vector<int> dp(n,0);
 6         if (n == 0)
 7             return 0;
 8         else if (n == 1 )
 9              return nums[0];
10         else if (n == 2)
11             return max(nums[0],nums[1]);
12         else
13             return max(robCore(nums,0,n-1),robCore(nums,1,n));
14         
15     }
16     int robCore (vector<int>& nums, int left, int right)
17     {
18         vector<int> dp(nums.size(),0);
19         dp[left] = nums[left];
20         dp[left+1] = max(nums[left],nums[left+1]); 
21         for (int i = left+2; i < right; i++)
22             {
23                 dp[i] = max(dp[i-1],dp[i-2]+nums[i]);
24             }
25         return dp[right-1];
26         
27     }
28 };

当然通过观察我们可以发现，其实只需要用两个变量，交替更新，就可以得到最后的结果，因此这道题最后可以简化为：

 1 class Solution {
 2 public:
 3     int rob(vector<int>& nums) {
 4         int n = nums.size();
 5         if(n == 0) return 0;
 6         if(n ==1) return nums[0];
 7         return max(robber(nums, 0, n-2), robber(nums, 1, n-1));
 8     }
 9     int robber(vector<int>&nums, int begin, int end){
10         int pre = 0, cur = 0;
11         for(int i = begin; i<= end; i++){
12             int tmp = max(pre+nums[i], cur);
13             pre = cur;
14             cur = tmp;
15         }
16         return cur;
17     }
18 };