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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.
这道题是之前那道House Robber 打家劫舍的拓展，现在房子排成了一个圆圈，则如果抢了第一家，就不能抢最后一家，因为首尾相连了，所以第一家和最后一家只能抢其中的一家，或者都不抢，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。那我们只需参考之前的House Robber 打家劫舍中的解题方法，然后调用两边取较大值，代码如下：
解法一
// DP class Solution { public: int rob(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) { if (right - left <= 1) return nums[left]; vector<int> dp(right, 0); dp[left] = nums[left]; dp[left + 1] = max(nums[left], nums[left + 1]); for (int i = left + 2; i < right; ++i) { dp[i] = max(nums[i] + dp[i - 2], dp[i - 1]); } return dp.back(); } };
解法二
class Solution { public: int rob(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); } };
[LeetCode] House Robber II 打家劫舍之二
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原文地址：http://www.cnblogs.com/grandyang/p/4518674.html
