740. 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: 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].

先统计各个数出现的次数，然后就是选了一些数就不能选旁边的数，DFS可定OK，但是DP更高效

class Solution:    def deleteAndEarn(self, nums):        """        :type nums: List[int]        :rtype: int        """        if not nums:    return 0        dict = {}        for i in nums:            if i in dict:  dict[i]+=i            else: dict[i] = i        keys = sorted(dict)        dp = [[0,0]]*(len(keys))        for i in range(len(keys)):            if i==0:                 dp[i][0], dp[i][1] = 0, dict[keys[i]]            elif keys[i]==keys[i-1]+1:                dp[i][0], dp[i][1] = max(dp[i-1][1], dp[i-1][0]), dp[i-1][0]+dict[keys[i]]            else:                dp[i][0], dp[i][1] = max(dp[i-1][1], dp[i-1][0]), max(dp[i-1][1], dp[i-1][0])+dict[keys[i]]                return max(dp[len(keys)-1][0], dp[len(keys)-1][1])        s = Solution()print(s.deleteAndEarn([1,1,1,2,4,5,5,5,6]))print(s.deleteAndEarn([3, 4, 2]))print(s.deleteAndEarn([2, 2, 3, 3, 3, 4]))

其实在Python里面初始化

dp = [[0,0]]*(len(keys))

是错的，但是因为这个DP可以做空间压缩，所以不影响结果