740. Delete and Earn

最新推荐文章于 2022-11-10 10:32:51 发布

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最新推荐文章于 2022-11-10 10:32:51 发布

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分类专栏： leetcode 动态规划 leetcode不会做的算法 Leetcode

本文链接：https://blog.youkuaiyun.com/qq_36718317/article/details/79820918

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本文介绍了一个有趣的问题：如何通过删除数组中的整数并获得相应分数来最大化总分。具体而言，一旦选择并删除一个元素，所有与其相差1的元素也会被删除。文章详细解释了解决方案，包括使用动态规划来找出最优解。

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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].

题意：

给出一组数，选择一个就删除它相邻大小的数，求最大的被选择数之和。

思路：

将值相等的数归在一起，设置一个数组sum，比如有三个3，那么sum[3]=9,这样处理之后，因为取一个数，那么与它相等的都会被取。再找状态转移方程，因为取i的话，那么i-1就取不了了，所以dp[i]=max(dp[i-2]+sum[i],sum[i-1])，因为自底向上求，所以dp可以用sum代替。

代码：

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
         vector<int> sums(10001, 0);
        for (int num : nums) sums[num] += num;
        for (int i = 2; i < 10001; ++i) {
            sums[i] = max(sums[i - 1], sums[i - 2] + sums[i]);
        }
        return sums[10000];
    }
};