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.
Notice
This is an extension of House Robber.
Example
nums = [3,6,4], return 6
LeetCode上的原题，请参见我之前的博客House Robber II。
解法一：
class Solution { public: /** * @param nums: An array of non-negative integers. * return: The maximum amount of money you can rob tonight */ int houseRobber2(vector<int>& nums) { if (nums.size() <= 1) return nums.empty() ? 0 : nums[0]; vector<int> a = nums, b = nums; a.erase(a.begin()); b.pop_back(); return max(rob(a), rob(b)); } int rob(vector<int> &nums) { if (nums.size() <= 1) return nums.empty() ? 0 : nums[0]; vector<int> dp{nums[0], max(nums[0], nums[1])}; for (int i = 2; i < nums.size(); ++i) { dp.push_back(max(dp[i - 2] + nums[i], dp[i - 1])); } return dp.back(); } };
解法二：
class Solution { public: /** * @param nums: An array of non-negative integers. * return: The maximum amount of money you can rob tonight */ int houseRobber2(vector<int>& nums) { if (nums.size() <= 1) return nums.empty() ? 0 : nums[0]; return max(rob(nums, 0, nums.size() - 1), rob(nums, 1, nums.size())); } int rob(vector<int> &nums, int left, int right) { int a = 0, b = 0; for (int i = left; i < right; ++i) { int m = a, n = b; a = n + nums[i]; b = max(m, n); } return max(a, b); } };
本文转自博客园Grandyang的博客，原文链接：打家劫舍之二[LintCode] House Robber II ，如需转载请自行联系原博主。
