House Robber II

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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.

Analyse: we should consider if the first house can be robbed. If we rob the first house, then the last house cannot be robbed. Instead, if we do not rob the first house, then we can rob the last house. Do the DP process two times and return the larger one.

Runtime: 0ms.

 1 class Solution {
 2 public:
 3     int rob(vector<int>& nums) {
 4         if(nums.size() == 0) return 0;
 5         if(nums.size() == 1) return nums[0];
 6         
 7         return max(helper(nums, 0, nums.size() - 1),
 8                    helper(nums, 1, nums.size()));
 9     }
10     int helper(vector<int>& nums, int start, int end){
11         int result = nums[start], dp1 = 0, dp2 = 0;
12         for(int i = start + 1; i < end; i++){
13             dp2 = dp1; //after computing, original dp1 becomes dp2
14             dp1 = result; //original result becomes dp1
15             result = max(dp1, dp2 + nums[i]);
16         }
17         return result;
18     }
19 };

House Robber II

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原文地址：http://www.cnblogs.com/amazingzoe/p/4743168.html