为有牺牲多壮志,敢教日月换新天。

[Swift]LeetCode1006. 笨阶乘 | Clumsy Factorial

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:山青咏芝(shanqingyongzhi)
➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址: https://www.cnblogs.com/strengthen/p/10504875.html 
➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎,请直接点击!!!

进入博主App Store主页,下载使用各个作品!!!

注:博主将坚持每月上线一个新app!!!

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.

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.

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.

Additionally, the division that we use is floor division such that 10 * 9 / 8 equals 11.  This guarantees the result is an integer.

Implement the clumsy function as defined above: given an integer N, it returns the clumsy factorial of N

Example 1:

Input: 4
Output: 7
Explanation: 7 = 4 * 3 / 2 + 1

Example 2:

Input: 10
Output: 12
Explanation: 12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1 

Note:

  1. 1 <= N <= 10000
  2. -2^31 <= answer <= 2^31 - 1  (The answer is guaranteed to fit within a 32-bit integer.)

通常,正整数 n 的阶乘是所有小于或等于 n 的正整数的乘积。例如,factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

相反,我们设计了一个笨阶乘 clumsy:在整数的递减序列中,我们以一个固定顺序的操作符序列来依次替换原有的乘法操作符:乘法(*),除法(/),加法(+)和减法(-)。

例如,clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1。然而,这些运算仍然使用通常的算术运算顺序:我们在任何加、减步骤之前执行所有的乘法和除法步骤,并且按从左到右处理乘法和除法步骤。

另外,我们使用的除法是地板除法(floor division),所以 10 * 9 / 8 等于 11。这保证结果是一个整数。

实现上面定义的笨函数:给定一个整数 N,它返回 N 的笨阶乘。 

示例 1:

输入:4
输出:7
解释:7 = 4 * 3 / 2 + 1

示例 2:

输入:10
输出:12
解释:12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1 

提示:

  1. 1 <= N <= 10000
  2. -2^31 <= answer <= 2^31 - 1  (答案保证符合 32 位整数。)

Runtime: 4 ms
Memory Usage: 18.9 MB
 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 }

 

posted @ 2019-03-10 12:19  为敢技术  阅读(384)  评论(0编辑  收藏  举报