[Swift]LeetCode509.斐波那契数|FibonacciNumber

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The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0and 1. That is,

F(0) = 0,   F(1) = 1
F(N) = F(N - 1) + F(N - 2), for N > 1.

Given N, calculate F(N).

Example 1:

Input: 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.

Example 2:

Input: 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.

Example 3:

Input: 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.

Note:

0 ≤ N ≤ 30.

斐波那契数，通常用 F(n) 表示，形成的序列称为斐波那契数列。该数列由 0 和 1 开始，后面的每一项数字都是前面两项数字的和。也就是：

F(0) = 0,   F(1) = 1
F(N) = F(N - 1) + F(N - 2), 其中 N > 1.

给定 N，计算 F(N)。

示例 1：

输入：2
输出：1
解释：F(2) = F(1) + F(0) = 1 + 0 = 1.

示例 2：

输入：3
输出：2
解释：F(3) = F(2) + F(1) = 1 + 1 = 2.

示例 3：

输入：4
输出：3
解释：F(4) = F(3) + F(2) = 2 + 1 = 3.

提示：

0 ≤ N ≤ 30

Runtime: 8 ms

Memory Usage: 18.3 MB

 1 class Solution {
 2     // f(0) = 0
 3     // f(1) = 1
 4     func fib(_ N: Int) -> Int {
 5         guard N > 1 else { return N }
 6         return processor(N, last: 1, llast: 0, anchor: 2)
 7     }
 8     
 9     func processor(_ n: Int, last: Int, llast: Int, anchor: Int) -> Int {
10         guard anchor < n else { return last + llast }
11         let value = last + llast
12         return processor(n, last: value, llast: last, anchor: anchor+1)
13     }
14 }

8ms

 1 class Solution {
 2     var cache : [Int:Int] = [:]
 3     
 4     func fib(_ N: Int) -> Int {
 5        if let fibAnswer = cache[N] {
 6            return fibAnswer
 7        }
 8         
 9         if N < 2{
10             return N
11         }
12         
13         let answer = fib(N - 1) + fib(N - 2)
14         cache[N] = answer
15         return answer
16     }
17 }

12ms

 1 class Solution {
 2     func fib(_ N: Int) -> Int {
 3         if N <= 1 { return N }
 4         
 5         var f = [Int](repeating: 0, count: N + 1)
 6         f[0] = 0
 7         f[1] = 1
 8         
 9         for i in 2...N {
10             f[i] = f[i - 2] + f[i - 1]
11         }
12         
13         return f[N]
14     }
15 }

16ms

 1 class Solution {
 2     func fib(_ N: Int) -> Int {
 3         if N < 2 {
 4             return N
 5         }
 6         
 7         var preValue = 1
 8         var prePreValue = 0
 9         var loop = 1
10         var result = 0
11         
12         while loop < N {
13             result = preValue + prePreValue
14             prePreValue = preValue
15             preValue = result
16             loop += 1
17         }
18         
19         return result
20     }
21 }

24ms

 1 class Solution {
 2 func fib(_ N: Int) -> Int {
 3     let ss = sqrt(5)
 4     let n = N 
 5     return Int((asdfasd((ss + 1) / 2, n) - asdfasd((1 - ss) / 2, n)) / ss)
 6 }
 7 
 8 func asdfasd(_ x: Double, _ N: Int) -> Double {
 9     var result: Double = 1
10     for _ in 0..<N {
11         result = result * x
12     }
13     return result
14   }
15 }

32ms

1 class Solution {
2     func fib(_ N: Int) -> Int {
3         if N == 0 || N == 1 {
4             return N
5         }
6         
7         return fib(N-1) + fib(N-2)
8     }
9 }