leetcode -- 740. Delete and Earn

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最新推荐文章于 2024-10-01 18:52:56 发布

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本文解析了LeetCode中一道题目“删除并获取积分”的解决方案，通过动态规划算法，详细介绍了如何在给定整数数组中选择元素进行删除以获取最大积分，避免选择相邻数值元素。

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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].

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/delete-and-earn
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路：

1、相邻的数字不能用

2、dp[i]记录到数字i时能获得的最大点数。要么删除这个数字，要么不删除这个数，删除上个数

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