【Leetcode】740. Delete and Earn
阿新 • • 發佈:2019-01-29
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
利用動態規劃的思想,先用計數器(counter)算出nums中每個數的數量,得到一個dic他, 然後從最小數開始遍歷:
1、當有兩個相差為1的值相鄰時,如([2,2,3]),計算2×2和3的最大值,存到dict[2]和dict[3]中
2、當有兩個相差大於1的值相鄰時,如([2,2,4]),則計算2×2+4,存到dict[4]中。
程式碼
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())
第一版程式碼是直接在counter得到的字典裡進行的操作。
第二版開闢了一個新陣列來存掙到的pointers,快了一丟丟,大約5%的樣子
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)