剑指offer：斐波那契数列

Posted 2021-02-21 smartheadliu

1. 递归

过不了！超时！

 1 class Solution {
 2 public:
 3     int Fibonacci(int n) {
 4         if(n==0)
 5             return 0;
 6         if(n==1)
 7             return 1;
 8         if(n==2)
 9             return 1;
10         if(n>2)
11             return Fibonacci(n-1)+Fibonacci(n-2);
12     }    
13 };

2. 递归改循环

 1 class Solution {
 2 public:
 3     int Fibonacci(int n) {
 4         if(n==0)
 5             return 0;
 6         if(n==1)
 7             return 1;
 8         if(n==2)
 9             return 1;
10         int n1=1,n2=1,res;
11         for(int i=3;i<=n;i++)
12         {
13             res=n1+n2;
14             n1=n2;
15             n2=res;
16         }
17         return res;
18     }
19 };