【leetcode】【198】House Robber

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#java #leetcode #dynamic programing

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本文探讨了一个经典的动态规划问题——打家劫舍。通过分析不同情况下的最优解策略，给出了详细的Java实现代码，帮助读者理解如何在不触动报警的情况下获取最大金额。

一、问题描述

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that 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.

二、问题分析

典型的dynamic programing。这种类型的题目，往往需要从头开始判断即初始状态，然后找出状态转移方程。结合我们的题目，当只有一间房子的时候，肯定偷；有两间房子的时候，选择一个钱多的偷；当有三间房子的时候，很明显要么偷第一间和第三间，要么偷第二间；依次类推。对于某间房子，只有两种状态：偷或者不偷。绕过弯来了，就好弄了。

三、Java AC代码

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