LeetCode in Kotlin
509. Fibonacci Number
Easy
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: n = 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
Example 2:
Input: n = 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
Example 3:
Input: n = 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
Constraints:
0 <= n <= 30
Solution
class Solution {
private val memo = IntArray(31)
fun fib(n: Int): Int {
if (n == 0) {
return 0
}
if (n == 1) {
return 1
}
if (memo[n] != 0) {
return memo[n]
}
memo[n] = fib(n - 1) + fib(n - 2)
return memo[n]
}
}