From f21fa6cd8ef0b262388d5479bfb272da5d14e869 Mon Sep 17 00:00:00 2001
From: chandan23sde <114254861+chandan23sde@users.noreply.github.com>
Date: Fri, 7 Oct 2022 19:39:13 +0530
Subject: [PATCH] =?UTF-8?q?Update=2010.1=20=E6=96=90=E6=B3=A2=E9=82=A3?=
 =?UTF-8?q?=E5=A5=91=E6=95=B0=E5=88=97.md?=
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I have added a simplified code of Time complexity O(n)
---
 ...2\243\345\245\221\346\225\260\345\210\227.md" | 16 ++++++++++++++++
 1 file changed, 16 insertions(+)

diff --git "a/notes/10.1 \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260\345\210\227.md" "b/notes/10.1 \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260\345\210\227.md"
index fd9e836bc1..a936422b15 100644
--- "a/notes/10.1 \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260\345\210\227.md"	
+++ "b/notes/10.1 \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260\345\210\227.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)。