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.

从一串数组中取数字，每取一个数num，所有的num+1和num-1都不能再取，求能取到的最大和。

解决：因为num取或不取，只跟num-1和num+1有关，所以想到的是先排序。并且只要num取了，num-1和num+1都不能取，所以num要么不取，要么所有的num都取。

因此，可以用一个map记录数据出现的次数。然后从头开始取数字。用earn表示取了这个数的和，nearn表示不取。

要是当前数字刚好只比前一个数字大了1，显然 earn[i] = nearn[i-1] + val[i]; 

要是当前数字比前一个数字大得多（>1），显然 earn[i] = max(earn[i-1], nearn[i-1]) + val[i];

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        map<int, int> m;
        for (auto num: nums)
            ++m[num];
        int earn = 0;
        int nearn = 0;
        int num = -1;
        map<int, int>::const_iterator it = m.begin();
        for (it = m.begin(); it!=m.end(); ++it) {
            if (it->first - 1 > num) {
                nearn = max(earn, nearn);
                earn = nearn + it->first * it->second;
            }
                
            else {
                int temp = nearn;
                nearn = max(earn, nearn);
                earn = temp + it->first * it->second;
            }
            num = it->first;
        }
        return max(earn, nearn);
    }
};