740. Delete and Earn

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: 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].

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;    }};