Python3解leetcode Maximum SubarrayHouse Robber

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.

 

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.

 

思路:

动态规划问题。从头开始遍历每一个house，判断如果rob当前house和不rob当前house这两种情况下，能够获得的最大收益各是多少，并且记录下来；基于第i个house的结果记录，可以推断出第i+1个house的结果记录

代码：

 

class Solution(object):
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) == 0 :#如果要是长度为0 ，则返回0
            return 0
        result = [nums[0],0]#result[0]代表rob nums[0]的结果，result[1]代表不rob  nums[0]的结果
        for i in range(1,len(nums)):#循环遍历，每次更新是以及否rob当前house的结果
            #account_include_me = nums[i] + result[1]
            # account_exclude_me = max(result)
            #result = [account_include_me,account_exclude_me]
            result = [nums[i] + result[1],max(result)]
        return max(result)

 

转载于:https://www.cnblogs.com/xiaohua92/p/11127109.html