★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號:山青詠芝(shanqingyongzhi)
➤博客園地址:山青詠芝(https://www.cnblogs.com/strengthen/)
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址: http://www.javashuo.com/article/p-vxojesfl-md.html
➤若是連接不是山青詠芝的博客園地址,則多是爬取做者的文章。
➤原文已修改更新!強烈建議點擊原文地址閱讀!支持做者!支持原創!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★html
Normally, the factorial of a positive integer n
is the product of all positive integers less than or equal to n
. For example, factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
.git
We instead make a clumsy factorial: using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.github
For example, clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
. However, these operations are still applied using the usual order of operations of arithmetic: we do all multiplication and division steps before any addition or subtraction steps, and multiplication and division steps are processed left to right.微信
Additionally, the division that we use is floor division such that 10 * 9 / 8
equals 11
. This guarantees the result is an integer.app
Implement the clumsy
function as defined above: given an integer N
, it returns the clumsy factorial of N
. less
Example 1:ide
Input: 4 Output: 7 Explanation: 7 = 4 * 3 / 2 + 1
Example 2:函數
Input: 10
Output: 12 Explanation: 12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
Note:this
1 <= N <= 10000
-2^31 <= answer <= 2^31 - 1
(The answer is guaranteed to fit within a 32-bit integer.)一般,正整數 n
的階乘是全部小於或等於 n
的正整數的乘積。例如,factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
。spa
相反,咱們設計了一個笨階乘 clumsy
:在整數的遞減序列中,咱們以一個固定順序的操做符序列來依次替換原有的乘法操做符:乘法(*),除法(/),加法(+)和減法(-)。
例如,clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
。然而,這些運算仍然使用一般的算術運算順序:咱們在任何加、減步驟以前執行全部的乘法和除法步驟,而且按從左到右處理乘法和除法步驟。
另外,咱們使用的除法是地板除法(floor division),因此 10 * 9 / 8
等於 11
。這保證結果是一個整數。
實現上面定義的笨函數:給定一個整數 N
,它返回 N
的笨階乘。
示例 1:
輸入:4 輸出:7 解釋:7 = 4 * 3 / 2 + 1
示例 2:
輸入:10 輸出:12 解釋:12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
提示:
1 <= N <= 10000
-2^31 <= answer <= 2^31 - 1
(答案保證符合 32 位整數。)1 class Solution { 2 func clumsy(_ N: Int) -> Int 3 { 4 if N == 1 {return 1} 5 if N == 2 {return 2} 6 if N == 3 {return 3 * 2 / 1} 7 return N * (N - 1) / (N - 2) + clumsyInternal(N - 3) 8 } 9 10 func clumsyInternal(_ N: Int) -> Int { 11 if N == 0 {return 0} 12 if N == 1 {return 1} 13 if N == 2 {return 1} 14 if N == 3 {return 3 - 2 * 1} 15 var ans:Int = N - (N - 1) * (N - 2) / (N - 3) 16 return ans + clumsyInternal(N - 4) 17 } 18 }
4ms
1 class Solution { 2 func clumsy(_ n: Int) -> Int { 3 if n == 1 { 4 return 1 5 } else if n == 2 { 6 return 2 7 } else if n == 3 { 8 return 6 9 } 10 var result = n*(n-1)/(n-2) 11 var n = n-3 12 while n >= 4 { 13 result += n - (n-1)*(n-2)/(n-3) 14 n -= 4 15 } 16 if n > 0 { 17 result += 1 18 } 19 return result 20 } 21 }
4ms
1 class Solution { 2 func clumsy(_ n: Int) -> Int { 3 switch n { 4 case 1: 5 return 1 6 case 2: 7 return 2 8 case 3: 9 return 6 10 default: 11 return n*(n-1)/(n-2) + clumsyInternal(n-3) 12 } 13 } 14 15 func clumsyInternal(_ n : Int ) -> Int { 16 if n >= 4 { 17 return n - (n-1)*(n-2)/(n-3) + clumsyInternal(n-4) 18 } 19 if n == 0 { 20 return 0 21 } 22 return 1 23 } 24 }
112ms
1 class Solution { 2 enum Operation: Int { 3 case mul 4 case div 5 case add 6 case sub 7 8 func next() -> Operation { 9 return Operation(rawValue: (self.rawValue + 1) % 4 )! 10 } 11 } 12 13 struct OperationCluster { 14 var current: Operation { 15 return operations[currentIndex] 16 } 17 18 var currentIndex: Int 19 20 let operations: [Operation] 21 init(operations: [Operation]) { 22 self.operations = operations 23 self.currentIndex = 0 24 } 25 26 mutating func next() { 27 currentIndex = (currentIndex + 1) % operations.count 28 } 29 } 30 31 func clumsy(_ N: Int) -> Int { 32 var a = [Int]() 33 34 for i in 0..<N { 35 a.append(N-i) 36 } 37 38 var multiplied = [Int]() 39 40 var operationCluster = OperationCluster(operations: [.mul, .div, .add, .sub]) 41 var ignoreNextOperand = false 42 43 for (index, number) in a.enumerated() { 44 defer { 45 operationCluster.next() 46 } 47 48 guard ignoreNextOperand == false else { 49 ignoreNextOperand = false 50 continue 51 } 52 53 switch operationCluster.current { 54 case .mul: 55 let result = number * (next(in: a, after: index) ?? 1) 56 multiplied.append(result) 57 ignoreNextOperand = true 58 case .div: 59 multiplied.append(number) 60 case .add: 61 multiplied.append(number) 62 case .sub: 63 multiplied.append(number) 64 } 65 } 66 67 operationCluster = OperationCluster(operations: [.div, .add, .sub]) 68 ignoreNextOperand = false 69 70 var divided = [Int]() 71 72 for (index, number) in multiplied.enumerated() { 73 defer { 74 operationCluster.next() 75 } 76 77 guard ignoreNextOperand == false else { 78 ignoreNextOperand = false 79 continue 80 } 81 82 switch operationCluster.current { 83 case .mul: 84 assertionFailure() 85 break 86 case .div: 87 let result = number / (next(in: multiplied, after: index) ?? 1) 88 divided.append(result) 89 ignoreNextOperand = true 90 case .add: 91 divided.append(number) 92 case .sub: 93 divided.append(number) 94 } 95 } 96 97 operationCluster = OperationCluster(operations: [.add, .sub]) 98 ignoreNextOperand = false 99 100 var signApplied = [Int]() 101 102 signApplied.append(divided.first ?? 0) 103 104 for (index, number) in divided.enumerated() { 105 defer { 106 operationCluster.next() 107 } 108 109 guard ignoreNextOperand == false else { 110 ignoreNextOperand = false 111 continue 112 } 113 114 switch operationCluster.current { 115 case .mul: 116 assertionFailure() 117 break 118 case .div: 119 assertionFailure() 120 break 121 case .add: 122 signApplied.append(next(in: divided, after: index) ?? 0) 123 case .sub: 124 signApplied.append(-(next(in: divided, after: index) ?? 0)) 125 } 126 } 127 128 return signApplied.reduce(0, +) 129 } 130 131 func next(in array: [Int], after index: Int) -> Int? { 132 let maxIndex = array.count - 1 133 if maxIndex <= index { 134 return nil 135 } 136 return array[index + 1] 137 } 138 }