[Swift]LeetCode1137. 第 N 个泰波那契数 | 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 定义如下： 

T0 = 0, T1 = 1, T2 = 1, 且在 n >= 0 的条件下 Tn+3 = Tn + Tn+1 + Tn+2

给你整数 n，请返回第 n 个泰波那契数 Tn 的值。 

示例 1：

输入：n = 4
输出：4
解释：
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4

示例 2：

输入：n = 25
输出：1389537 

提示：

• 0 <= n <= 37
• 答案保证是一个 32 位整数，即 answer <= 2^31 - 1。

---

0ms

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

---

4ms

class Solution {
    func tribonacci(_ n: Int) -> Int {
        if n == 0 {
            return 0
        } else if n == 1 {
            return 1
        } else if n == 2 {
            return 1
        }
        var a = 0
        var b = 1
        var c = 1
        var d = 0
        for i in 3...n {
            d = a + b + c
            a = b
            b = c
            c = d
        }
        return d
    }
}

---

8ms

class Solution {
    func tribonacci(_ n: Int) -> Int {
        var tribs = [Int](repeating: 0, count: max(n + 1, 3))
        tribs[0] = 0
        tribs[1] = 1
        tribs[2] = 1
        guard n >= 3 else {
            return tribs[n]
        }
        for i in 3...n {
            tribs[i] = tribs[i - 1] + tribs[i - 2] + tribs[i - 3] 
        }
        return tribs[n]
    }
}

---

Runtime: 12 ms

Memory Usage: 20.8 MB

class Solution {
    func tribonacci(_ n: Int) -> Int {
        var first:Int = 0
        var second:Int = 1
        var third:Int = 1
        // here first,second and third are the previous three elements
        switch n
        {
            case 0:
            return 0
            case 1,2:
            return 1
            default:
            // if n=0 or n=1 then the tribonacci number is n itself 
            // else for loop will be executed
            for i in 3...n
            {
                var newElement:Int = first + second + third
                first=second
                second=third
                third=newElement
            }
        }
        return third        
    }
}