diff --git a/notes/10.1 斐波那契数列.md b/notes/10.1 斐波那契数列.md
index fd9e836b..a936422b 100644
--- a/notes/10.1 斐波那契数列.md	
+++ b/notes/10.1 斐波那契数列.md	
@@ -31,6 +31,22 @@ public int Fibonacci(int n) {
     return fib[n];
 }
 ```
+OR Simplified code
+class FibonacciExample1{  
+public static void main(String args[])  
+{    
+ int n1=0,n2=1,n3,i,count=10;    
+ System.out.print(n1+" "+n2);//printing 0 and 1    
+    
+ for(i=2;i<count;++i)//loop starts from 2 because 0 and 1 are already printed    
+ {    
+  n3=n1+n2;    
+  System.out.print(" "+n3);    
+  n1=n2;    
+  n2=n3;    
+ }    
+  
+}}  
 
 考虑到第 i 项只与第 i-1 和第 i-2 项有关，因此只需要存储前两项的值就能求解第 i 项，从而将空间复杂度由 O(N) 降低为 O(1)。