740. Delete and get points

The comment said it was a house robbery 4, It's very figurative .
The meaning of this question is to delete a number to obtain points , Then force the deletion i+1 and i-1 And you can't get points , We need to get the maximum number of points .
Let's go through nums, Record the number of occurrences of each number , And record the largest number .
It mainly depends on whether to take the current figure , If you take the front, you can't take , If you don't take the front, you can take ,0 I just didn't take it ,1 It's just taking
State definition ：dp[i][1] Is to delete the current number ,dp[i][0] Just don't delete the current number
State shift : If you take the front, you can't take （ The previous ones are forcibly deleted ）, If you don't take the front, you can take it instead , Take it, then =dp[i-1][0] + ima[i], Use the current number Number of occurrences
Finally back to dp[maxn][0] and [1] The bigger one

class Solution {
    
public:
    int deleteAndEarn(vector<int>& nums) {
    
        int n = nums.size();
        int maxn = 0;
        unordered_map<int,int> ma;
        for(int i = 0; i < n; ++i){
    
            ma[nums[i]] += 1;
            maxn = max(maxn,nums[i]);
        }
        vector<vector<int>> dp(maxn+1,vector<int>(2,0));
        // It mainly depends on whether to take the current figure , If you take the front, you can't take , If you don't take the front, you can take 
        //0 I just didn't take it ,1 It's just taking 
        for(int i = 1; i <= maxn; ++i){
    
            dp[i][1] = dp[i-1][0] + i * ma[i];
            dp[i][0] = max(dp[i-1][1],dp[i-1][0]);
        }
        return max(dp[maxn][0],dp[maxn][1]);
    }
};