LeetCode之Fibonacci Number（Kotlin）

問題： 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 0 and 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.
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方法： 初始化一個list，然後每次新增的元素等於倒數第一個和倒數第二個元素的和，最後輸出list中的第N個元素即為最終結果。

具體實現：

class FibonacciNumber {
    fun fib(N: Int): Int {
        val list = mutableListOf(0, 1)
        for (i in 2..N) {
            list.add(list[i - 1] + list[i - 2])
        }
        return list[N]
    }
}

fun main(args: Array<String>) {
    val num = 2
    val fibonacciNumber = FibonacciNumber()
    println(fibonacciNumber.fib(num))
}
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有問題隨時溝通

具體程式碼實現可以參考Github