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.
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
与House Robber的差异在于，nums[0]和nums[n-1]不能同时包含，
因此等同于nums[0...n-2]和nums[1...n-1]两者间取较大值。
class Solution { public: int rob(vector<int>& nums) { if(nums.empty()) return 0; if(nums.size() == 1) return nums[0]; vector<int> nums1(nums); vector<int> nums2(nums); nums1.erase(nums1.begin()); nums2.pop_back(); return max(originRob(nums1), originRob(nums2)); } int originRob(vector<int>& nums) { if(nums.empty()) return 0; int prev2 = 0; int prev1 = nums[0]; for(int i = 1; i < nums.size(); i ++) { int cur = max(prev2+nums[i], prev1); prev2 = prev1; prev1 = cur; } return prev1; } };
