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509. 斐波那契数
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,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
}