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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.
Runtime: 0ms
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 if(n == 2) return max(nums[0], nums[1]); 8 9 int *dp1 = new int[n]; 10 dp1[0] = nums[0]; 11 dp1[1] = nums[0]; 12 //rub the first house 13 for(int i = 2; i < n - 1; i++){ 14 dp1[i] = max(dp1[i -2] + nums[i], dp1[i - 1]); 15 } 16 dp1[n - 1] = dp1[n - 2]; 17 18 //do not rub the first house 19 int *dp2 = new int[n]; 20 dp2[0] = 0; 21 dp2[1] = nums[1]; 22 for(int i = 2; i < n; i++){ 23 dp2[i] = max(dp2[i - 2] + nums[i], dp2[i - 1]); 24 } 25 26 return max(dp1[n - 1], dp2[n -1]); 27 } 28 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/4809043.html
