LeetCode-House Robber II-解题报告_house robber二 leetcodr-优快云博客

本文探讨了LeetCode上环形房屋抢劫者II问题的解决方案，该问题是原房屋抢劫者问题的一个扩展。文章介绍了一种通过两次动态规划来解决首位相连的环形布局问题的方法。

原题链接：https://leetcode.com/problems/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.

和house robber不同的地方在于首位也算相邻，只需要将环状dp变成线性dp

1 - n dp一次。

0 - n-1 dp一次。

取两次两次dp的最大值

我开始没想清楚，用了3次dp，不清不楚就ac了。。。

class Solution {
public:
    int rob(vector<int>& nums) {
		int n = nums.size();
		if (n == 0)return 0;
		if (n == 1)return nums[0];
		if (n == 2)return max(nums[0], nums[1]);
		if (n == 3)return max(max(nums[0], nums[1]), nums[2]);
		int prepre = nums[0], t;
		int pre = nums[1];
		pre = max(prepre, pre);
		int ans;
		for (int i = 2; i < n - 1; ++i)
		{
			t = pre;
			pre = max(prepre + nums[i], pre);
			prepre = t;
		}
		ans = pre;
		prepre = nums[1];
		pre = nums[2];
		pre = max(prepre, pre);
		for (int i = 3; i < n; ++i)
		{
			t = pre;
			pre = max(prepre + nums[i], pre);
			prepre = t;
		}
		ans = max(pre, ans);

		prepre = nums[2];
		pre = nums[3];
		pre = max(prepre, pre);
		for (int i = 4; i < n; ++i)
		{
			t = pre;
			pre = max(prepre + nums[i], pre);
			prepre = t;

		}
		ans = max(pre, ans);
		return ans;
	}
};