本文主要是介绍[LeetCode] 198. House Robber，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目内容
https://leetcode-cn.com/problems/house-robber/
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.
Example 1: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.
Example 2:Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).Total amount you can rob = 2 + 9 + 1 = 12.题目思路
这道题目不难，就是可以用动态规划的方法即可。设置一个动态规划的数组dp，dp[i]表示在确认入侵nums[i]的情况下，可以最多的收获是多少。
程序代码
class Solution(object):def rob(self, nums):""":type nums: List[int]:rtype: int"""l=len(nums)if l==0:return 0if l==1:return nums[0]if l==2:return max(nums[0],nums[1])if l==3:return max(nums[1],nums[0]+nums[2])dp=[0]*ldp[0],dp[1],dp[2]=nums[0],nums[1],nums[0]+nums[2]for i in range(3,l):dp[i]=nums[i]+max(dp[i-2],dp[i-3])return max(dp)
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