剑指offer 10-I. 斐波那契数列

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题目描述

写一个函数,输入 n ,求斐波那契(Fibonacci)数列的第 n 项。斐波那契数列的定义如下:

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

斐波那契数列由 0 和 1 开始,之后的斐波那契数就是由之前的两数相加而得出。

答案需要取模 1e9+7(1000000007),如计算初始结果为:1000000008,请返回 1。

示例 1:

输入:n = 2
输出:1

示例 2:

输入:n = 5
输出:5

提示:

  • 0 <= n <= 100 解法

递推求解

Python3

class Solution:
    def fib(self, n: int) -> int:
        a, b = 0, 1
        for _ in range(n):
            a, b = b, a + b
        return a % 1000000007

Java

class Solution {
    public int fib(int n) {
        int a = 0, b = 1;
        for (int i = 0; i < n; ++i) {
            int c = (a + b) % 1000000007;
            a = b;
            b = c;
        }
        return a;
    }
}

C++

class Solution {
public:
    int fib(int n) {
        int a = 0, b = 1;
        for (int i = 0; i < n; ++i) {
            int c = (a + b) % 1000000007;
            a = b;
            b = c;
        }
        return a;
    }
};

JavaScript

/**
 * @param {number} n
 * @return {number}
 */
var fib = function (n) {
  let a = 0,
    b = 1;
  for (let i = 0; i < n; ++i) {
    const c = (a + b) % (1e9 + 7);
    a = b;
    b = c;
  }
  return a;
};