740. Delete and Earn

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本文介绍了一种名为DeleteandEarn的算法问题，该问题要求在遵循特定规则的情况下从整数数组中删除元素以获得最大得分。文章详细解释了解决方案的步骤，并提供了C++实现代码。通过将数组转化为值数组，简化了问题并使用动态规划的方法求解。

740. Delete and Earn

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].

Analyze:
首先，题目表示取了nums[i]的值，那么nums[i-1]和nums[i+1]的值都不能选了，那也就是只要出现nums[i]的值，不论我取多少次，对于nums[i-1]和nums[i+1]的影响都是一样的，那么为了最大化我的盈利，我会取尽nums[i]。根据这个性质，我们可以将nums转化成对应的values，即将nums[i]出现的次数乘nums[i]，即得到values[nums[i]]的值。
其次，考虑到取了nums[i]的值，那么nums[i-1]和nums[i+1]的值这个特性，可以推断出数字是不能连续取的，那么就可以将状态简化成以下两个状态：

take[i] = skip[i-1] + value[i];
skip[i] = max(skip[i-1], take[i-1]);

Code:

#include<iostream>
#include<vector>

using namespace std;

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        int value[10001] = { 0 };
        for (int i = 0; i < nums.size(); i++)
            value[nums[i]] += nums[i];

        int take = 0, skip = 0;
        for (int i = 0; i < 10001; i++) {
            int takei = skip + value[i];
            int skipi = skip > take ? skip : take;
            take = takei;
            skip = skipi;
        }
        return skip > take ? skip : take;
    }
};