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 = [2,2,3,4] 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
nums
is at most20000
. - Each element
nums[i]
is an integer in the range[1,10000]
.
给定一个整数数组 nums
,你可以对它进行一些操作。
每次操作中,选择任意一个 nums[i]
,删除它并获得 nums[i]
的点数。之后,你必须删除每个等于 nums[i] - 1
或 nums[i] + 1
的元素。
开始你拥有 0 个点数。返回你能通过这些操作获得的最大点数。
示例 1:
输入: nums = [3,2] 输出: 6 解释: 删除 4 来获得 4 个点数,因此 3 也被删除。 之后,删除 2 来获得 2 个点数。总共获得 6 个点数。
示例 2:
输入: nums = [2,4] 输出: 9 解释: 删除 3 来获得 3 个点数,接着要删除两个 2 和 4 。 之后,再次删除 3 获得 3 个点数,再次删除 3 获得 3 个点数。 总共获得 9 个点数。
注意:
-
nums
的长度最大为20000
。 - 每个整数
nums[i]
的大小都在[1,10000]
范围内。
36ms
1 class Solution { 2 func deleteAndEarn(_ nums: [Int]) -> Int { 3 if nums.count == 0 { 4 return 0 5 } 6 7 let maxN = nums.max()! 8 var M = [Int](repeating: 0,count: maxN + 1) 9 var DP = [Int](repeating: 0,count: maxN + 1) 10 11 for i in 0..<nums.count { 12 M[nums[i]] += nums[i] 13 } 14 15 DP[0] = M[0] 16 DP[1] = M[1] 17 for i in 2..<M.count { 18 DP[i] = max(DP[i - 2] + M[i],DP[i - 1]) 19 } 20 21 return DP.last! 22 } 23 }
40ms
1 class Solution { 2 func deleteAndEarn(_ nums: [Int]) -> Int { 3 let minValue = nums.min() ?? 0 4 let maxValue = nums.max() ?? 0 5 var dp = [Int](repeating: 0,count:maxValue + 1) 6 for item in nums { 7 dp[item] += item 8 } 9 10 if maxValue >= 2{ 11 for i in 2...maxValue { 12 dp[i] = max(dp[i-2] + dp[i],dp[i-1]) 13 } 14 } 15 16 return dp[maxValue] 17 } 18 }
44ms
1 class Solution { 2 func deleteAndEarn(_ nums: [Int]) -> Int { 3 var counts = [Int: Int]() 4 for num in nums { 5 counts[num,default: 0] += 1 6 } 7 8 var prev = -1 9 var avoid = 0 10 var using = 0 11 12 for num in counts.keys.sorted() { 13 if num - 1 != prev { 14 (avoid,using) = (max(avoid,using),num * counts[num]! + max(avoid,using)) 15 } else { 16 (avoid,num * counts[num]! + avoid) 17 } 18 19 prev = num 20 } 21 22 return max(avoid,using) 23 } 24 }
Runtime: 52 ms
Memory Usage: 19.3 MB
1 class Solution { 2 func deleteAndEarn(_ nums: [Int]) -> Int { 3 var sums:[Int] = [Int](repeating:0,count:10001) 4 for num in nums 5 { 6 sums[num] += num 7 } 8 for i in 2..<10001 9 { 10 sums[i] = max(sums[i - 1],sums[i - 2] + sums[i]) 11 } 12 return sums[10000] 13 } 14 }
92ms
1 class Solution { 2 func deleteAndEarn(_ nums: [Int]) -> Int { 3 if nums.count == 0 { 4 return 0 5 } 6 if nums.count == 1 { 7 return nums[0] 8 } 9 var numCount: [String: Int] = [:] 10 var dp: [String : Int] = [:] 11 var maxCount = 0 12 for i in 0 ..< nums.count { 13 if let a = numCount["\(nums[i])"] { 14 numCount["\(nums[i])"] = a + 1 15 }else { 16 numCount["\(nums[i])"] = 1 17 } 18 if nums[i] > maxCount { 19 maxCount = nums[i] 20 } 21 } 22 dp["0"] = 0 23 if let a = numCount["1"] { 24 dp["1"] = a 25 }else { 26 dp["1"] = 0 27 } 28 for i in 2 ... maxCount { 29 var a = 0 30 if let e = dp["\(i - 1)"] { 31 a = e 32 } 33 var b = 0 34 var c = 0 35 if let count = dp["\(i - 2)"] { 36 c = count 37 } 38 if let count = numCount["\(i)"] { 39 b = count * i + c 40 } 41 dp["\(i)"] = max(a,b) 42 } 43 return dp["\(maxCount)"]! 44 } 45 }
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。