斐波那契数列

• 问题：数列第n项的值

• 抽象状态F(i)：数列第i项的值

• 转移方程：F(i) : F(i-1)+F(i-2)

• 初始状态：我们转移方程需要俩项，才能运行起来，

  1. F(0) = 0
  2. F(1) = 1
  3. 返回F(n)

第一种：递归

public class Solution {
    
    
    public int Fibonacci(int n) {
    
    
        if(n == 0)return 0;
        if(n == 1 || n == 2)return 1;
        int f3 = Fibonacci(n-1)+Fibonacci(n-2);
        return f3;
    }
}

第二种:非递归

public class TestDemo1 {
    
    
    public static void main(String[] args) {
    
    
        //斐波那契非递归(输出第n项的值)
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        System.out.println(Fibonacci(n));
    }
    public static int Fibonacci(int n) {
    
    
        if(n == 0)return 0;
        if(n == 1 || n == 2)return 1;
        int[] arr = new int[n];
        arr[0] = 1;
        arr[1] = 1;
        for (int i = 2;i < n;i++){
    
    
            arr[i] = arr[i-1]+arr[i-2];
        }
        return arr[n-1];
    }
 }