斐波那契数

题目

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).

提示:

• 0 <= n <= 30

示例

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

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

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

题解

JavaScript - 动态规划一维数组

/**
 * @param {number} n
 * @return {number}
 */
var fib = function (n) {
  if (n < 2) return n

  const dp = new Array(n + 1).fill(0)
  dp[0] = 0
  dp[1] = 1

  for (let i = 2; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2]
  }

  return dp[n]
}

JavaScript - 动态规划优化版

/**
 * @param {number} n
 * @return {number}
 */
var fib = function (n) {
  if (n < 2) return n

  let prev = 0
  let curr = 1

  for (let i = 2; i <= n; i++) {
    const sum = prev + curr
    prev = curr
    curr = sum
  }

  return curr
}