[LeetCode] 509. Fibonacci Number-优快云博客

本文链接：https://blog.youkuaiyun.com/liuhk333/article/details/104036500

本文深入解析了斐波那契数列的计算方法，对比了递归与非递归算法，详细阐述了如何使用非递归方式高效求解斐波那契数列，提供了一个C++实现的示例。

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Descriptions:

The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,

F(0) = 0, F(1) = 1
F(N) = F(N - 1) + F(N - 2), for N > 1.
Given N, calculate F(N).

Example 1:

Input: 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
Example 2:

Input: 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
Example 3:

Input: 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.

Note:

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。

Analysis:

大一上学期刚学编程就已经做过的题，当时印象是学递归编程的例题。从此以后见到这道题就想起递归解法。

递归解法是比较好理解并且容易书写，然而非递归的方式也是可以很好地解决这一道题目，每一次计算，等式中的3个参数就右移一位。要获取F(n)，只需循环计算到相应次数就行了。

时间复杂度O(n)，空间复杂度O(1)

C++ solution:

class Solution {
public:
    int fib(int N) {
        if(N == 0) return 0;
        else if(N == 1) return 1;
        else 
        {
            int r_num = 0;
            int mid_num = 1;
            int l_num = mid_num + r_num;
            for(size_t n = 2; n < N; ++n)
            {
                r_num = mid_num;
                mid_num = l_num;
                l_num += r_num;
            }
            return l_num;
        }
    }
};