How to delete and get points in LeetCode
Editor to share with you how to delete LeetCode and get points, I believe most people do not know much about it, so share this article for your reference, I hope you can learn a lot after reading this article, let's go to know it!
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Topic description
Given an integer array nums, select any nums [I] in each operation, delete it and get the points of nums [I], and delete each element equal to nums [I]-1 or nums [I] + 1 synchronously. The initial number of points is 0, which returns the maximum number of points that can be obtained. For example: nums= [3Jing 4jue 2], return 6. (first select 4, accumulate 4 points, delete 3 at the same time, select 2, and then accumulate 2 points, for a total of 6 points. The points accumulated in other ways are less than 6)
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Answer to the question
Idea: dynamic planning through the title requirements, the first thing to be clear is: when you select a value to get its points, other same values will also be selected. Because when you select a value, the adjacent values have been deleted, so the other same values will not be deleted and must be selected. So you can create a list value with values as the subscript to record the number of points you can get when you select each value. For example, value [2] = 2, 2], [3] = 3, [4] = 4: 1, so we finally get value= [0].
Intermediate state: where the intermediate state dp [I] represents the maximum number of points that can be obtained from 1 selection to I.
State transition: there are only two possibilities for a value to be deleted and the number of points to be obtained, and whether it is obtained or not will affect the selection of other values. Therefore, if a new number is not selected, then DP [I] = DP [I-1]; if elected, iMui 1 cannot be selected, then DP [I] = DP [I-2] + value [I], so eventually dp [I] = max (DP [I-1], DP [I-2] + value [I]).
Class Solution: def deleteAndEarn (self, nums: List [int])-> int: if not nums: return 0 dp = [0] * (max (nums) + 1) value = dp.copy () # value = [0] * (max (nums) + 1) for an in enumerate (nums): value [a] + = a dp [1] = value [1] for i in range (2) Max (nums) + 1): dp [I] = max (DP [I-1], DP [I-2] + value [I]) return dp [- 1] above are all the contents of the article "how to delete and get points by LeetCode" Thank you for reading! I believe we all have a certain understanding, hope to share the content to help you, if you want to learn more knowledge, welcome to follow the industry information channel!
