[LeetCode] 213. House Robber II

轉自LeetCode

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.

<Solution>

198. House Robber 的衍生題

想法如下

• 將搶第一間房子和搶最後一間房子分開來看，分開來看之後，就不用考慮 circle arrangement 的情況

• 用之前的 dp 解法分別計算出各自的最佳解後，再找出最大值

code 如下

Java

	class Solution {
	public int rob(int[] nums) {
	if(nums.length <= 1) {
	return nums.length == 0 ? 0 : nums[0];
	}
	//>> divided to two cases
	//>> 1. rob the first house (last house cannot be robbed)
	//>> 2. rob the last house (first house cannot be robbed)
	//>> each case use dp solution to solve without considering the circle arrangement
	return Math.max(robHouse(nums, 0, nums.length-1), robHouse(nums, 1, nums.length));
	}
	private int robHouse(int[] nums, final int start, final int end) {
	if(end - start <= 1) {
	return nums[start];
	}
	int[] dp = new int[end];
	dp[start] = nums[start];
	dp[start+1] = Math.max(nums[start], nums[start+1]);
	for(int i = start+2; i < end; i++) {
	dp[i] = Math.max(nums[i] + dp[i-2], dp[i-1]);
	}
	return dp[end-1];
	}
	}

view raw rob2.java hosted with ❤ by GitHub

Kotlin

	class Solution {
	fun rob(nums: IntArray): Int {
	return when(nums.size) {
	1 -> nums[0]
	2 -> Math.max(nums[0], nums[1])
	else -> {
	var dp = IntArray(nums.size) { 0 }
	dp[0] = nums[0]
	dp[1] = Math.max(nums[0], nums[1])
	//>> rob house 1
	for(i in 2..nums.lastIndex-1) {
	dp[i] = Math.max(dp[i-1], dp[i-2] + nums[i])
	}
	val profit1 = dp[nums.lastIndex-1]
	//>> without robbing house 1
	dp[0] = 0
	dp[1] = nums[1]
	for(i in 2..nums.lastIndex) {
	dp[i] = Math.max(dp[i-1], dp[i-2] + nums[i])
	}
	Math.max(profit1, dp[nums.lastIndex])
	}
	}
	}
	}

view raw house_robber_II.kt hosted with ❤ by GitHub