斐波那契数实现

时间复杂度O(n)

 1 long Fibonacci( long n )
 2 {
 3     if( n <= 1 )
 4     return n;
 5     
 6     long one=0;
 7     long two=1;
 8     
 9     for( long i = 2; i <= n; ++i )
10     {
11         int tmp = two;
12         two += one;
13         one = tmp;
14     }
15     return two;
16 }

 

posted @ 2013-07-19 09:52  NinaGood  阅读(144)  评论(0)  编辑  收藏  举报