740. Delete and Earn

最新推荐文章于 2025-06-04 07:15:00 发布

转载最新推荐文章于 2025-06-04 07:15:00 发布 · 52 阅读

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原文链接：http://www.cnblogs.com/iamxiaoyubei/p/8278273.html

本文探讨了一种针对整数数组的操作方式，通过删除特定元素并获取相应分数的方法，同时必须移除相邻数值的元素。提供了详细的算法实现，旨在帮助读者理解如何获得执行此类操作可能得到的最大总分数。

#week14

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: 6
Explanation: 
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: 9
Explanation: 
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].

题解：

 1 class Solution {
 2 public:
 3     int deleteAndEarn(vector<int>& nums) {
 4         int n = 10001;
 5         vector<int> values(n, 0);
 6         for (int num : nums)
 7             values[num] += num;
 8 
 9         int take = 0, skip = 0;
10         for (int i = 0; i < n; i++) {
11             int takei = skip + values[i];
12             int skipi = max(skip, take);
13             take = takei;
14             skip = skipi;
15         }
16         return max(take, skip);
17     }
18 };

转载于:https://www.cnblogs.com/iamxiaoyubei/p/8278273.html