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 ofnums
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]; } };