LeetCode[213] House Robber II

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#leetcode #动态规划

解决环形排列房屋的抢劫问题，避免抢相邻房屋并确保不被警察发现。将问题拆分为两部分，利用动态规划算法求解最大抢劫数额。

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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.

不能抢相邻的房子，nums开头和结尾相邻，不能同时抢，于是可以分到两个不同的数组里。然后每个数组用动态规划，状态方程为

dp[i] = max(dp[i - 1], dp[i - 2] + nums[i + start]);

class Solution {
public:
	int rob(vector<int>& nums) {
		if (nums.size() <= 1)
			return nums.empty() ? 0 : nums[0];
		return max(HouseRob(nums, 0, nums.size() - 2), HouseRob(nums, 1, nums.size() - 1));
	}
	int HouseRob(vector<int>& nums, int start, int end) {
		if (start == end)
			return nums[start];
		vector<int> dp(nums.size() - 1, 0);
		dp[0] = nums[start];
		dp[1] = max(nums[start], nums[start + 1]);
		for (int i = 2; i < nums.size() - 1; ++i)
		{
			dp[i] = max(dp[i - 1], dp[i - 2] + nums[i + start]);
		}
		return dp[nums.size() - 2];
	}
};