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:

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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:

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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].

动态规划问题，对于nums中出现的最大的数n来说，有两种可能：

  1. 能挣到点数，那么最大的分数就是n×n出现的次数+最大值为n-2时的分数
  2. 不能挣到点数，那么最大值就是最大值为n-1的分数

因为nums的元素取值范围为[1,10000]所以可以用哈希表来保存每个数字出现了多少次。

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

        vector<int> dp(10001, 0);
        dp[1] = flag[1];
        dp[2] = max(flag[2] * 2, dp[1]);
        for (int i = 3; i < dp.size(); i++) {
            dp[i] = max(dp[i - 1], dp[i - 2] + flag[i] * i);
        }
        return dp.back();
    }
};