编程开发习题Delete and Earn

编程开发习题Delete and Earn。

原题目如下所示：

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].

题目的大概意思就是给出一组数，然后你需要通过删除其中的一个数并获得其数值相等的点数，假设这个数为i，那么一旦选了i，就要从数组中删除与i-1或i+1数值相等的数，求能得到的最大点数。

这道题开始入手时就想到应该是一个动态规划问题，可以从子问题来开始考虑动态规划的方程，可以用skip和take两个数组来分别代表当前最大点数或是选择这个数时的点数，即出现下列动态规划方程：

take[i] = skip[i-1] + values[i];
skip[i] = Math.max(skip[i-1], take[i-1]);

所以根据上述方程就可以进行编程了，源代码如下：

class Solution {
public:
    int deleteAndEarn(vector& nums) {
        int n = 10001;
        vector values(n, 0);
        for (int num : nums)
            values[num] += num;

        int take = 0, skip = 0;
        for (int i = 0; i < n; i++) {
            int takei = skip + values[i];
            int skipi = max(skip, take);
            take = takei;
            skip = skipi;
        }
        return max(take, skip);
    }
};