[Swift]LeetCode509. 斐波那契数 | Fibonacci Number

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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

class Solution {
    // f(0) = 0
    // f(1) = 1
    func fib(_ N: Int) -> Int {
        guard N > 1 else { return N }
        return processor(N, last: 1, llast: 0, anchor: 2)
    }
    
    func processor(_ n: Int, last: Int, llast: Int, anchor: Int) -> Int {
        guard anchor < n else { return last + llast }
        let value = last + llast
        return processor(n, last: value, llast: last, anchor: anchor+1)
    }
}

---

8ms

class Solution {
    var cache : [Int:Int] = [:]
    
    func fib(_ N: Int) -> Int {
       if let fibAnswer = cache[N] {
           return fibAnswer
       }
        
        if N < 2{
            return N
        }
        
        let answer = fib(N - 1) + fib(N - 2)
        cache[N] = answer
        return answer
    }
}

---

12ms

class Solution {
    func fib(_ N: Int) -> Int {
        if N <= 1 { return N }
        
        var f = [Int](repeating: 0, count: N + 1)
        f[0] = 0
        f[1] = 1
        
        for i in 2...N {
            f[i] = f[i - 2] + f[i - 1]
        }
        
        return f[N]
    }
}

---

16ms

class Solution {
    func fib(_ N: Int) -> Int {
        if N < 2 {
            return N
        }
        
        var preValue = 1
        var prePreValue = 0
        var loop = 1
        var result = 0
        
        while loop < N {
            result = preValue + prePreValue
            prePreValue = preValue
            preValue = result
            loop += 1
        }
        
        return result
    }
}

---

24ms

class Solution {
func fib(_ N: Int) -> Int {
    let ss = sqrt(5)
    let n = N 
    return Int((asdfasd((ss + 1) / 2, n) - asdfasd((1 - ss) / 2, n)) / ss)
}

func asdfasd(_ x: Double, _ N: Int) -> Double {
    var result: Double = 1
    for _ in 0..<N {
        result = result * x
    }
    return result
  }
}

---

32ms

class Solution {
    func fib(_ N: Int) -> Int {
        if N == 0 || N == 1 {
            return N
        }
        
        return fib(N-1) + fib(N-2)
    }
}