House Robber II

最新推荐文章于 2025-08-14 09:11:10 发布

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最新推荐文章于 2025-08-14 09:11:10 发布

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分类专栏： leetcode 文章标签： java 动态规划 leetcode

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解决环形排列房屋的抢劫问题，确保首尾房屋不会同时被抢。采用动态规划方法，分两部分求解并返回最大值。

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题目描述：

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.

先写上House Robber.的解答吧，太简单了，就全部一起写了：

public class Solution {
    public int rob(int[] nums) {
		if(nums.length==0)
			return 0;
		if(nums.length==1)
        	return nums[0];
		if(nums.length==2)
			return Math.max(nums[0], nums[1]);
		int[] dp=new int[nums.length];
		dp[0]=nums[0];
		dp[1]=Math.max(nums[0], nums[1]);
		for(int i=2;i<nums.length;i++){
			dp[i]=Math.max(dp[i-1], dp[i-2]+nums[i]);
		}
		return dp[nums.length-1];
    }
}

而本题不允许第一个和最后一个一起被偷，那么我们可以假设小偷从nums[0]，偷到nums[n-2]，然后又从nums[1]偷到nums[n-1]，这样就可以使得nums[0]和nums[n-1]永远不会一起被偷了，然后比较他们的最大值，就是被偷的最大值。

代码如下：

public class Solution {
    public int rob(int[] nums) {
		if(nums.length==0)
			return 0;
		if(nums.length==1)
        	return nums[0];
		if(nums.length==2)
			return Math.max(nums[0], nums[1]);
		if(nums.length==3)
			return Math.max(Math.max(nums[0], nums[1]),nums[2]);
		
		int[] dp1=new int[nums.length-1];
		dp1[0]=nums[0];
		dp1[1]=Math.max(nums[0], nums[1]);
		for(int i=2;i<nums.length-1;i++){
			dp1[i]=Math.max(dp1[i-1], dp1[i-2]+nums[i]);
		}
		
		int[] dp2=new int[nums.length-1];
		dp2[0]=nums[1];
		dp2[1]=Math.max(nums[1], nums[2]);
		for(int i=3;i<nums.length;i++){
			dp2[i-1]=Math.max(dp2[i-2], dp2[i-3]+nums[i]);
		}
		
		return Math.max(dp1[dp1.length-1], dp2[dp2.length-1]);
    }
}