LeetCode之Fibonacci Number（Kotlin

问题：
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.

方法：
初始化一个list，然后每次新增的元素等于倒数第一个和倒数第二个元素的和，最后输出list中的第N个元素即为最终结果。

具体实现：

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))
}

有问题随时沟通

具体代码实现可以参考Github