[Swift]LeetCode1137. 第 N 個泰波那契數 | N-th Tribonacci Number

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The Tribonacci sequence Tn is defined as follows: 

T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0.

Given n, return the value of Tn. 

Example 1:

Input: n = 4
Output: 4
Explanation:
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4

Example 2:

Input: n = 25
Output: 1389537 

Constraints:

• 0 <= n <= 37
• The answer is guaranteed to fit within a 32-bit integer, ie. answer <= 2^31 - 1.

---

泰波那契序列 Tn 定義以下： 

T0 = 0, T1 = 1, T2 = 1, 且在 n >= 0 的條件下 Tn+3 = Tn + Tn+1 + Tn+2

給你整數 n，請返回第 n 個泰波那契數 Tn 的值。 

示例 1：

輸入：n = 4
輸出：4
解釋：
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4

示例 2：

輸入：n = 25
輸出：1389537 

提示：

• 0 <= n <= 37
• 答案保證是一個 32 位整數，即 answer <= 2^31 - 1。

---

0ms

 1 class Solution {
 2     func tribonacci(_ n: Int) -> Int {
 3         var numbers = Array<Int>(repeating: 0, count: 38)
 4         numbers[1] = 1
 5         numbers[2] = 1
 6         for i in 3..<38 {
 7             numbers[i] = numbers[i-1] + numbers[i-2] + numbers[i-3]
 8         }
 9         return numbers[n]
10     }
11 }

---

4ms

 1 class Solution {
 2     func tribonacci(_ n: Int) -> Int {
 3         if n == 0 {
 4             return 0
 5         } else if n == 1 {
 6             return 1
 7         } else if n == 2 {
 8             return 1
 9         }
10         var a = 0
11         var b = 1
12         var c = 1
13         var d = 0
14         for i in 3...n {
15             d = a + b + c
16             a = b
17             b = c
18             c = d
19         }
20         return d
21     }
22 }

---

8ms

 1 class Solution {
 2     func tribonacci(_ n: Int) -> Int {
 3         var tribs = [Int](repeating: 0, count: max(n + 1, 3))
 4         tribs[0] = 0
 5         tribs[1] = 1
 6         tribs[2] = 1
 7         guard n >= 3 else {
 8             return tribs[n]
 9         }
10         for i in 3...n {
11             tribs[i] = tribs[i - 1] + tribs[i - 2] + tribs[i - 3] 
12         }
13         return tribs[n]
14     }
15 }

---

Runtime: 12 ms

Memory Usage: 20.8 MB

 1 class Solution {
 2     func tribonacci(_ n: Int) -> Int {
 3         var first:Int = 0
 4         var second:Int = 1
 5         var third:Int = 1
 6         // here first,second and third are the previous three elements
 7         switch n
 8         {
 9             case 0:
10             return 0
11             case 1,2:
12             return 1
13             default:
14             // if n=0 or n=1 then the tribonacci number is n itself 
15             // else for loop will be executed
16             for i in 3...n
17             {
18                 var newElement:Int = first + second + third
19                 first=second
20                 second=third
21                 third=newElement
22             }
23         }
24         return third        
25     }
26 }