[LeetCode] 740. Delete and Earn

轉自LeetCode

You are given an integer array nums. You want to maximize the number of points you get by performing the following operation any number of times:

• Pick any nums[i] and delete it to earn nums[i] points. Afterwards, you must delete every element equal to nums[i] - 1 and every element equal to nums[i] + 1.

Return the maximum number of points you can earn by applying the above operation some number of times.

 

Example 1:

Input: nums = [3,4,2]
Output: 6
Explanation: You can perform the following operations:
- Delete 4 to earn 4 points. Consequently, 3 is also deleted. nums = [2].
- Delete 2 to earn 2 points. nums = [].
You earn a total of 6 points.

Example 2:

Input: nums = [2,2,3,3,3,4]
Output: 9
Explanation: You can perform the following operations:
- Delete a 3 to earn 3 points. All 2's and 4's are also deleted. nums = [3,3].
- Delete a 3 again to earn 3 points. nums = [3].
- Delete a 3 once more to earn 3 points. nums = [].
You earn a total of 9 points.

 

Constraints:

• 1 <= nums.length <= 2 * 104
• 1 <= nums[i] <= 104

Solution

當刪除了 nums[i]，nums[i] - 1 和 nums[i] + 1 就不能使用

其實也就是，相鄰的值不能取，和 198. House Robber 是同樣的題目

那需要做些改變的是，必須是找相鄰的數字

那題目有說，數字是從 1 到 10000，所以用一個長度 10001 的 int array 來存

並把數字，以及相對應的分數給算起來

之後一樣用 DP 的概念去解就可以，max(從前一個來，從前兩個來 + 當前的值)

kotlin

	class Solution {
	fun deleteAndEarn(nums: IntArray): Int {
	val sum = IntArray(10001) {0}
	for(n in nums) {
	sum[n] += n
	}
	for(i in 2..10000) {
	sum[i] = Math.max(sum[i-1], sum[i-2] + sum[i])
	}
	return sum.last()
	}
	}

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