[LeetCode]740. Delete and Earn

• descrption
  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: 6Explanation: 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: 9Explanation: 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].
• 解题思路
  这道题大致意思是让随意挑选数字nums[i]删除它,得到nums[i]分，但是同时必须删除数组中所有等于nums[i]-1和nums[i]+1的元素，直到数组中的所有元素都被删除。这里用到动态规划求解，可以用数组times[]记录数组num[]中每个数字出现的次数，即times[i]就是数组num[]中i出现的次数。按照题目要求，如果我们选择数字i，那么相应的就不能选择i-1和i+1，这里可以定义状态转移方程：

用points[i]记录遍历到数字i时的最大得分 ：points[i] = max(points[i-2]+times[i]*i, points[i-1])

• 代码如下

class Solution {public:    int deleteAndEarn(vector<int>& nums) {        vector<int> times(10001, 0);        //count appear times for every num in nums        for (int i = 0; i < nums.size(); i++) {            times[nums[i]]++;        }        vector<int> points(10001, 0);        points[1] = times[1]*1;        for (int i = 2; i <= points.size(); i++) {            points[i] = max(points[i-1], points[i-2]+times[i]*i);        }        return points[10000];    }};

原题地址

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