数据结构 | Fibonacci数列的递归、数组、迭代(循环)实现及其时间复杂度

#include <stdio.h>
#include <iostream>

int fibonacci_recurrence(int n) //递归实现
{
    if (n == 0 or n == 1)
        return 1;
    else if(n > 1)
    {
        return fibonacci_recurrence(n-1)+fibonacci_recurrence(n-2);
    }
    else
        printf("error");
}//递归实现的时间复杂度是O(2^n)

int fibonacci_array(int n) //数组实现
{
    //1,1,2,3,5,8,13,......
    if (n == 0 or n == 1)
        return 1;
    else if (n > 1)
    {
        int nums[n];
        nums[0] = 1;
        nums[1] = 1;
        for(int i = 2; i<=n; i++)
        {
            nums[i] = nums[i-1] + nums[i-2];
        }
        return nums[n];
    }
}//数组实现的时间复杂度是O(n)

int fibonacci_iteration(int n) //迭代(循环)实现
{
    if (n == 0 or n == 1)
        return 1;
    else
    {
        int num1 = 1;
        int num2 = 1;
        int res;
        while(n > 1)
        {
            res = num1 + num2;
            num2 = num1;
            num1 = res;
            n--;
        }
        return res;
    }
}//迭代实现的时间复杂度是O(n)

int main()
{
    int num;
    printf("input num:\n");
    scanf("%d",&num);
    if(num<0)
        printf("error");
    else
    {
        printf("递归实现：\n");
        printf("%d\n",fibonacci_recurrence(num));
        printf("数组实现：\n");
        printf("%d\n",fibonacci_array(num));
        printf("迭代(循环)实现：\n");
        printf("%d\n",fibonacci_iteration(num));
    }
    system("pause");
    return 0;
}