Leetcode刷题:740. Delete and Earn

Given an array nums of integers, you can perform operations on the array.

In each operation, you pick any nums[i] and delete it to earn nums[i] points. After, you must delete every element equal to nums[i] - 1 or nums[i] + 1.

You start with 0 points. Return the maximum number of points you can earn by applying such operations.

思路

利用动态规划的思想,先用计数器(counter)算出nums中每个数的数量,得到一个dict, 然后从最小数开始遍历:
1、当有两个相差为1的值相邻时,如([2,2,3]),计算2*2和3的最大值,存到dict[2]和dict[3]中
2、当有两个相差大于1的值相邻时,如([2,2,4]),则计算2*2+4,存到dict[4]中。

代码

version 1.

class Solution(object):
    def deleteAndEarn(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        if len(nums) == 1:
            return nums[0]
        dict = collections.Counter(nums)
        sort_dict = sorted(dict.keys())
        dict[sort_dict[0]] = dict[sort_dict[0]] * sort_dict[0]
        if sort_dict[1] - sort_dict[0] == 1:
            dict[sort_dict[1]] = max(dict[sort_dict[1]] * sort_dict[1], dict[sort_dict[0]])
        else:
            dict[sort_dict[1]] = dict[sort_dict[1]] * sort_dict[1] + dict[sort_dict[0]]
        for i in range(2,len(sort_dict)):
            if sort_dict[i] - sort_dict[i-1] == 1:
                dict[sort_dict[i]] = max(dict[sort_dict[i]] * sort_dict[i] +dict[sort_dict[i-2]],
                                         dict[sort_dict[i-1]])
            else:
                dict[sort_dict[i]] = dict[sort_dict[i]] * sort_dict[i] +dict[sort_dict[i-1]]
        return max(dict.values())

第一版代码是直接在counter得到的字典里进行的操作。
第二版开辟了一个新数组来存挣到的pointers,快了一丢丢,大约5%的样子
version 2:

        if not nums:
            return 0
        if len(nums) == 1:
            return nums[0]
        dict = collections.Counter(nums)
        dp = [0] * (len(nums))
        sort_dict = sorted(dict.keys())
        dp[0] = dict[sort_dict[0]] * sort_dict[0]
        if sort_dict[1] - sort_dict[0] == 1:
            dp[1] = max(dict[sort_dict[1]] * sort_dict[1], dp[0])
        else:
            dp[1] = dict[sort_dict[1]] * sort_dict[1] + dp[0]
        for i in range(2,len(sort_dict)):
            if sort_dict[i] - sort_dict[i-1] == 1:
                dp[i] = max(dict[sort_dict[i]] * sort_dict[i] +dp[i-2],
                                         dp[i-1])
            else:
                dp[i] = dict[sort_dict[i]] * sort_dict[i] +dp[i-1]
        return max(dp)

猜你喜欢

转载自blog.csdn.net/a529975125/article/details/79712643