509. Fibonacci Number

Description

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

Solutions

Solution 1

Python Code
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class Solution:
    def fib(self, n: int) -> int:
        a, b = 0, 1
        for _ in range(n):
            a, b = b, a + b
        return a

Java Code
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class Solution {
    public int fib(int n) {
        int a = 0, b = 1;
        while (n-- > 0) {
            int c = a + b;
            a = b;
            b = c;
        }
        return a;
    }
}

C++ Code
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class Solution {
public:
    int fib(int n) {
        int a = 0, b = 1;
        while (n--) {
            int c = a + b;
            a = b;
            b = c;
        }
        return a;
    }
};

Go Code
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func fib(n int) int {
	a, b := 0, 1
	for i := 0; i < n; i++ {
		a, b = b, a+b
	}
	return a
}

TypeScript Code
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function fib(n: number): number {
    let a = 0;
    let b = 1;
    for (let i = 0; i < n; i++) {
        [a, b] = [b, a + b];
    }
    return a;
}

Rust Code
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impl Solution {
    pub fn fib(n: i32) -> i32 {
        let mut a = 0;
        let mut b = 1;
        for _ in 0..n {
            let t = b;
            b = a + b;
            a = t;
        }
        a
    }
}

JavaScript Code
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/**
 * @param {number} n
 * @return {number}
 */
var fib = function (n) {
    let a = 0;
    let b = 1;
    while (n--) {
        const c = a + b;
        a = b;
        b = c;
    }
    return a;
};

PHP Code
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class Solution {
    /**
     * @param Integer $n
     * @return Integer
     */
    function fib($n) {
        if ($n == 0 || $n == 1) {
            return $n;
        }
        $dp = [0, 1];
        for ($i = 2; $i <= $n; $i++) {
            $dp[$i] = $dp[$i - 2] + $dp[$i - 1];
        }
        return $dp[$n];
    }
}

Solution 2

TypeScript Code
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function fib(n: number): number {
    if (n < 2) {
        return n;
    }
    return fib(n - 1) + fib(n - 2);
}

Rust Code
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impl Solution {
    pub fn fib(n: i32) -> i32 {
        if n < 2 {
            return n;
        }
        Self::fib(n - 1) + Self::fib(n - 2)
    }
}