[LeetCode]Delete and Earn题解（动态规划）

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].

这是一题对我很有启发的动态规划。

解题思路：

先将数组的各个value放到另一个数组set中，其下标是value（和桶排序第一步一样），set中元素的值是index对应的数之和。（例如，数组1，3，3，4对应的桶，set[1] = 1,set[3] = 6,set[4] =4.）

设我们选中i元素时得到的分数是take[i]，不选该元素时得到的分数是skip[i]，动态规划的思路是：

第i个元素对应的take[i]是：

第i-1个元素不被选中的分数（也就是skip[i-1]）+第i个元素的值（此时第i个元素被选中） 和

第i-1个元素被选中的分数（也就是take[i-1]）（此时第i个元素不被选中）

之中最大的那个值。

第i个元素对应的skip[i]是：

第i-1个元素被选中的分数（也就是take[i-1]）

一直到最后得出的take就是我们要的值。

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        int take = 0, skip = 0;
        std::vector<int> set = (1001,0);
        if(nums.empty()) return 0;
        for(int i = 0; i < nums.size(); i++){
            set[nums[i]] += nums[i];//初始化
        }
        for(int i = 0; i < nums.size(); i++){
            /*
            skip+set[i]:前一个元素不选的分数+当前元素选中的分数
            take:前一个元素选中的分数（+当前元素不选的分数，也即是0）
            */
            int temp = max(skip + set[i], take);
            skip = take;//当前元素不选的分数 = 前一个元素选中的分数 + 0
            take = temp;//更新当前元素选中的分数
        }
        return take;
    }
};