Question

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

方法一

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

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;
  }
}

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

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

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;
}

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
  }
}

/**
 * @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;
};

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];
  }
}

方法二

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

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