213. House Robber II #

题目 #

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

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.

Example 1:

Input: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.

Example 2:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

题目大意 #

你是一个专业的小偷，计划偷窃沿街的房屋，每间房内都藏有一定的现金。这个地方所有的房屋都围成一圈，这意味着第一个房屋和最后一个房屋是紧挨着的。同时，相邻的房屋装有相互连通的防盗系统，如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警。

给定一个代表每个房屋存放金额的非负整数数组，计算你在不触动警报装置的情况下，能够偷窃到的最高金额。

解题思路 #

• 这一题是第 198 题的加强版。不过这次是在一个环形的街道中，即最后一个元素和第一个元素是邻居，在不触碰警报的情况下，问能够窃取的财产的最大值是多少？
• 解题思路和第 198 完全一致，只需要增加额外的一个转换。由于首尾是相邻的，所以在取了第一个房子以后就不能取第 n 个房子，那么就在 [0,n - 1] 的区间内找出总价值最多的解，然后再 [1,n] 的区间内找出总价值最多的解，两者取最大值即可。

代码 #

package leetcode

func rob213(nums []int) int {
	n := len(nums)
	if n == 0 {
		return 0
	}
	if n == 1 {
		return nums[0]
	}
	if n == 2 {
		return max(nums[0], nums[1])
	}
	// 由于首尾是相邻的，所以需要对比 [0，n-1]、[1，n] 这两个区间的最大值
	return max(rob213_1(nums, 0, n-2), rob213_1(nums, 1, n-1))
}

func rob213_1(nums []int, start, end int) int {
	preMax := nums[start]
	curMax := max(preMax, nums[start+1])
	for i := start + 2; i <= end; i++ {
		tmp := curMax
		curMax = max(curMax, nums[i]+preMax)
		preMax = tmp
	}
	return curMax
}