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509. 斐波那契数

Open buuing opened this issue 4 years ago • 0 comments

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 01 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,计算 F(N)

示例 1:

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

示例 2:

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

示例 3:

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

提示:

  • 0 ≤ N ≤ 30

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/fibonacci-number 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。




  • 暴力解法
const fib = n => {
  if (n === 0) return 0
  if (n === 1) return 1
  return fib(n - 1) + fib(n - 2)
}
  • 记忆化递归
const fib = n => {
  const memo = [0, 1]
  const dfs = (i) => {
    if (memo[i] === undefined) {
      memo[i] = dfs(i - 1) + dfs(i - 2)
    }
    return memo[i]
  }
  return dfs(n)
}
  • 动态规划
const fib = n => {
  const dp = [0, 1]
  for (let i = 2; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2]
  }
  return dp[n]
}
  • 动态规划-状态压缩
const fib = n => {
  if (n === 0) return 0
  if (n === 1) return 1
  let prev = 0, curr = 1, next
  for (let i = 2; i <= n; i++) {
    next = prev + curr
    prev = curr
    curr = next
  }
  return curr
}

buuing avatar Dec 24 '20 07:12 buuing