0509. Fibonacci Number
Easy | DP | 12 ms (96.11%), 13.3 MB (90.24%)
Source: LeetCode - Fibonacci Number GitHub: Solution / Performance
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).
Constraints:
0 <= n <= 30
Input: n = 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
Input: n = 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
Input: n = 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.It is quite straightforward that two variables store information of F(n-1) and F(n-2) respectively. And the initialized values are 0 and 1 respectively.
class Solution:
def fib(self, n: int) -> int:
# (base case)
if n == 0 or n == 1: return n
# ==================================================
# Dynamic Programming =
# ==================================================
# time : O(n)
# space : O(1)
n1, n2 = 0, 1
for i in range(n - 1):
n1, n2 = n2, n1 + n2
return n2
class Solution {
/**
* @time : O(n)
* @space : O(1)
*/
public int fib(int n) {
/* base case */
if(n == 0 || n == 1) return n;
int first = 0, second = 1;
for(int i=0 ; i<n-1 ; i++) {
int tmp = first;
first = second;
second += tmp;
}
return second;
}
}Last updated
Was this helpful?