> For the complete documentation index, see [llms.txt](https://hadyang.gitbook.io/interview/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hadyang.gitbook.io/interview/content/docs/offer/fibonacci/index.md).

# index

***

### title: 斐波纳切数列 date: 2019-08-21T11:00:41+08:00 draft: false categories: offer

### 题目

[牛客网](https://www.nowcoder.com/practice/c6c7742f5ba7442aada113136ddea0c3?tpId=13\&tqId=11160\&tPage=1\&rp=1\&ru=%2Fta%2Fcoding-interviews\&qru=%2Fta%2Fcoding-interviews%2Fquestion-ranking)

大家都知道斐波那契数列，现在要求输入一个整数n，请你输出斐波那契数列的第n项（从0开始，第0项为0）。`n<=39`

### 解题思路

1. 递归计算很慢，是最简单的算法

```
public int Fibonacci(int n) {
    if (n == 0) {
        return 0;
    }

    if (n == 1) {
        return 1;
    }

    int l = 1, ll = 0;

    for (int i = 2; i <= n; i++) {
        int t = ll + l;
        ll = l;
        l = t;
    }

    return l;
}
```