README.md

509. 斐波那契数

English Version

题目描述

斐波那契数 （通常用 F(n) 表示）形成的序列称为 斐波那契数列 。该数列由 0 和 1 开始，后面的每一项数字都是前面两项数字的和。也就是：

F(0) = 0，F(1) = 1
F(n) = F(n - 1) + F(n - 2)，其中 n > 1

给定 n ，请计算 F(n) 。

 

示例 1：

输入：n = 2
输出：1
解释：F(2) = F(1) + F(0) = 1 + 0 = 1

示例 2：

输入：n = 3
输出：2
解释：F(3) = F(2) + F(1) = 1 + 1 = 2

示例 3：

输入：n = 4
输出：3
解释：F(4) = F(3) + F(2) = 2 + 1 = 3

 

提示：

• 0 <= n <= 30

解法

Python3

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

Java

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++

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

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

JavaScript

/**
 * @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;
};

TypeScript

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;
}

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

Rust

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
    }
}

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

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