【CODE】Delete and Earn

740. Delete and Earn

Medium

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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].

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        map<int,int> mp;
        for(int i=0;i<nums.size();i++) mp[nums[i]]+=nums[i];
        int max1=0,max2=0;
        //max1:取当前数的最大和
        //max2:不取当前数的最大和
        map<int,int>::iterator it1,it2;
        for(it1=mp.begin();it1!=mp.end();it1++){
            if(it1==mp.begin()){
                max1=it1->second;
                max2=0;
            }else{
                int tmax1=max1,tmax2=max2;
                max2=max(tmax1,tmax2);//不取当前数的最大值=max(取前一个数的最大值，不取前一个数的最大值)
                if(it1->first==it2->first+1) max1=tmax2+it1->second;
                //it2是it1的前一个，map是有序的。因为上一轮末尾，it2=it1，而这一轮开头，it1++
                //如果it2与it1的键值相邻，即当前数与上一个数是相邻的，那么取当前数的最大值=不取上一个数的最大值+当前数值
                else max1=max2+it1->second;
                //如果it2与it1的键值不相邻，那么取当前数的最大值=不取当前数的最大值+当前数值
            }
            it2=it1;//it2要跟上
        }
        return max(max1,max2);
    }
};

• Runtime: 8 ms, faster than 79.76% of C++ online submissions for Delete and Earn.
• Memory Usage: 9.4 MB, less than 75.00% of C++ online submissions for Delete and Earn.
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