leetcode213 -- House Robber II

leetcode213 -- House Robber II

题目

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, 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: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.

Example 2:

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.


这个题目的意思是现在房子是围成一个圈了,这意味着第一个和最后一个不能够同时偷了,主要是如何把这个问题转化为之前熟悉的那个一条街上的。

现在想,如果可以考虑偷第一家的话,那么最后一家就一定不能偷,同样,如果考虑偷最后一家的话,那么第一家就不能够偷,所以最后取两种情况的最大值就可以了。


class Solution:
    def rob(self, nums: List[int]) -> int:
    
        if len(nums)==0:
            return 0
        if len(nums)==1:
            return nums[0]
        if len(nums)==2:
            return max(nums)
        
        return max(self.helper(nums[:-1]), self.helper(nums[1:]))
        
    def helper(self, nums):
        if len(nums)==0:
            return 0
        if len(nums) ==1:
            return nums[0]
    
        dp = [0]*len(nums)
        dp[0] = nums[0]
        dp[1] = max(nums[:2])
        
        for i in range(2, len(nums)):
            dp[i]=max(dp[i-1], dp[i-2]+nums[i])
            
        return dp[-1]

法二

这种法也比较好,就是分奇偶性

class Solution:
    def rob(self, nums: List[int]) -> int:
    
        if len(nums)==0:
            return 0
        if len(nums)==1:
            return nums[0]
        if len(nums)==2:
            return max(nums)
        
        return max(self.helper(nums[:-1]), self.helper(nums[1:]))
        
    def helper(self, nums):
        if len(nums)==0:
            return 0
        if len(nums) ==1:
            return nums[0]
    
        dp = [0]*len(nums)
        dp[0] = nums[0]
        dp[1] = max(nums[:2])
        
        for i in range(2, len(nums)):
            dp[i]=max(dp[i-1], dp[i-2]+nums[i])
            
        return dp[-1]

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