Algorithm and data structure surface question (17)-Fibonacci number and __java

Topics

Title: The definition Fibonacci sequence is as follows:
/0 n=0
f (n) = 1 n=1
\ f (n-1) +f (n-2) n=2
8
Enter N, the quickest way to find the nth item of the sequence.
requires time complexity of Olog (n)

Thinking of solving problems

Previous articles: http://zhedahht.blog.163.com/blog/static/25411174200722991933440/

the problem translates to the following formula

This 2x2 matrix is represented by a matrix object. There are 4 properties, representing F (n), F (n-1), F (n-1), F (n-2), and F (n) to get the value of the first property.

the formula of the matrix is obtained by the method of division and treatment.

This gives the value of the n-1 of the matrix {1,1,1,0}.

The following is a formula that multiplies 2 matrices:

Code

public class Problem19 {    //2x2 matrix     class matrix2x2 {      
 //first Row first position         int pos00;
        //First row second position         int pos01;
        //second row first position         int Pos10;

        //second row second position         int pos11;         public matrix2x2 () {        }     & nbsp;   public matrix2x2 (int pos00, int pos01, int pos10, int pos11) {        &nb
sp;   this.pos00 = pos00;
            this.pos01 = pos01;
            this.pos10 = POS10;
            this.pos11 = POS11;         }     }     //2 2x2 matrix multiply to get new matrix      Class Matrixmultiply {        public matrix2x2 Multiply (matrix2x2 M1, matrix2x2 m2) {&nbsp ;           return New matrix2x2 (M1.POS00 * m2.pos00 + m1.pos01 * M2.POS10,                      m1.pos00 * m2.pos01 + m1.pos01 * M2.POS11, M1.pos10                     & nbsp;       * m2.pos00 + m1.pos11 * m2.pos10, M1.POS10         &NB sp;                   * m2.pos01 + m1.pos11 *
M2.POS11);         }     }     matrixmultiply multiplyer = new MATRIXMU

Ltiply ();     // Fibonacci number Nth number     public matrix2x2 getfibo (int n) {        if (n < 0) {            return null;         }  & nbsp
     matrix2x2 matrix22 = null;         if (n = = 1) {            matrix22 = NE
W matrix2x2 (1, 1, 1, 0);        &nbsp,}         //even        & Nbsp;else if (n% 2 = = 0) {            matrix22 = Getfibo (N/2);  &nbs P
         matrix22 = multiplyer.multiply (matrix22, matrix22);         }         //odd        & Nbsp;else if (n% 2! = 0) {            //extracts the first matrix, gives birthThe matrix is an even number, but keep in mind that the first matrix to be extracted is counted at the end.
            matrix22 = Getfibo ((n-1)/2);
            matrix22 = multiplyer.multiply (matrix22, matrix22);             matrix22 = multiplyer.multiply (matrix22, New matrix2x2 (1, 1, 1
, 0));
        }         return matrix22;     }     //original recursion     public static int GetFibo2 (int n) {   &nbsp ;    {            if (n==0| | n==1)             {          
      return n;            }             else             {                return GetFibo2 (n-1) +getfibo2 (n-2)
;            }         }      }          public static void Main (string[] args) {        in
T n = 30;
        matrix2x2 matrix = new Problem19 (). Getfibo (n-1);
        int fn = matrix.pos00;
        system.out.println (FN);
        int fn2 = Problem19.getfibo2 (n);
        system.out.println (FN2);
     }}

Output

832040
832040

Extended

Some netizens say can be solved by the following formula

Fibonacci General formula:

an=1/√[(1+√5/2) n (1-√5/2) n]