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.

Note:

The length of  nums is at most  20000.

Each element  nums[i] is an integer in the range  [1, 10000].

题意：

给出一组数，选择一个就删除它相邻大小的数，求最大的被选择数之和。

思路：

将值相等的数归在一起，设置一个数组sum，比如有三个3，那么sum[3]=9,这样处理之后，因为取一个数，那么与它相等的都会被取。再找状态转移方程，因为取i的话，那么i-1就取不了了，所以dp[i]=max(dp[i-2]+sum[i],sum[i-1])，因为自底向上求，所以dp可以用sum代替。

代码：

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
         vector<int> sums(10001, 0);
        for (int num : nums) sums[num] += num;
        for (int i = 2; i < 10001; ++i) {
            sums[i] = max(sums[i - 1], sums[i - 2] + sums[i]);
        }
        return sums[10000];
    }
};