198.House Robber（dp）

标签：exp   out   ash   ice   constrain   integer   determine   ESS   sys   

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.

class Solution:
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        dp = [0 for i in range(len(nums))]
        if len(nums)==0:
            return 0
        if len(nums)==1:
            return nums[0]
        dp[0] = nums[0]
        dp[1] = max(nums[0],nums[1])
        for i in range(2,len(nums)):
            if dp[i-1]==dp[i-2]:
                dp[i] = dp[i-1] + nums[i]
            else:
                dp[i] = max(dp[i-1],dp[i-2]+nums[i])
        return dp[-1]

如果dp的思路不好想的话，可以尝试从第一个开始，多写几个找到规律，写状态转移方程。

198.House Robber（dp）

标签：exp   out   ash   ice   constrain   integer   determine   ESS   sys   

原文地址：https://www.cnblogs.com/bernieloveslife/p/9768963.html