Delete and Earn
Question
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
numsis at most20000. - Each element
nums[i]is an integer in the range[1, 10000].
Answer
这个问题有两个点：
- 解与问题的输入数组顺序无关。
- 这是一个动态规划问题。
我们可以通过将每一种数字的值累加起来，得到每一种数字的总和。比如：
[2, 2, 3, 3, 3, 4] -> [0, 0, 4, 9, 4]即将对应的0,1,2,3,4的数字的和加起来赋值到对应的下标。
代码如下
class Solution(object):
def deleteAndEarn(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
numSum = [0]*10001
dp = [0] * 10001
for num in nums:
numSum[num] += num
for i in range(1, len(dp)):
if i == 1: dp[i] = numSum[i]
else: dp[i] = max(dp[i-1], dp[i-2]+numSum[i])
return max(dp)
