leetcode198 -- House Robber I

leetcode198 -- House Robber I

题目



u 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.




解答

class Solution:
    def rob(self, nums: List[int]) -> int:
        
        # 用dp[i]表示经过第i个房子之后的最大利润,那么有两种情况
        
        # case1. 当前的房子偷了,dp[i-2]+nums[i]
        # case2 .当前的房子没有偷 dp[i-1]
        
        # 所以dp[i]= max(dp[i-1], dp[i-2]+nums[i])
        
        if len(nums)==0:
            return 0
        if len(nums) ==1:
            return nums[0]
        # next len(nums) >=2
        dp = [0]*len(nums)
        dp[0] = nums[0]
        dp[1] = max(nums[0], nums[1])
        
        for i in range(2, len(nums)):
            dp[i]= max(dp[i-1], dp[i-2]+nums[i])
        
        return dp[-1]

注意

这个题有个地方很容易出现误解, 即并不是只有两种情况,即一种是全偷奇数的,另一种是全偷偶数的, 比如说[2,1,1,2] 最多可以偷到4.

打赏,谢谢~~

取消

感谢您的支持,我会继续努力的!

扫码支持
扫码打赏,多谢支持~

打开微信扫一扫,即可进行扫码打赏哦