[leetcode] 740. Delete and Earn

[leetcode] 740. Delete and Earn

Description

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.

Example 1:

Input: nums = [3, 4, 2]Output: 6Explanation: Delete 4 to earn 4 points, consequently 3 is also deleted.Then, delete 2 to earn 2 points. 6 total points are earned.

Example 2:

Input: nums = [2, 2, 3, 3, 3, 4]Output: 9Explanation: Delete 3 to earn 3 points, deleting both 2's and the 4.Then, delete 3 again to earn 3 points, and 3 again to earn 3 points.9 total points are earned.

Note:

The length of nums is at most 20000.Each element nums[i] is an integer in the range [1, 10000].

分析

题目的意思是：给你一个数组，每删除一个数，你就会获得那个位置的数值，但是同时你要删一个那个数-1的数，或者该数值+1的数，然后求你最终能够获得的最大的值。

我们首先统计每个数的求和，然后就可以转化成house robber的问题了。dp[i]在于拿或者不拿当前的数得到的最大的数值。这道题我做不出来，看来能够正确理解题意，对解决问题的帮助很大。

代码

class Solution {public: int deleteAndEarn(vector& nums) { vector sums(10001,0); for(auto num:nums){ sums[num]+=num; } vector dp(sums.size(),0); dp[1]=sums[1]; for(int i=2;i

参考文献

​​Delete and Earn问题及解法​​​​[LeetCode] Delete and Earn 删除与赚取​​

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