时间复杂度为O( n )的方法：

 

long long Fibonacci( unsigned n )   {         int result[2] = {
   0, 1};         if(n < 2)               return result[n];         long long   fibOne = 0;         long long   fibTwo = 1;         long long   fibThree ;             for(unsigned int i = 2; i <= n; ++ i)          {                fibThree = fibOne + fibTwo;     　　　　fibOne = fibTwo ;                fibNTwo = fibThree;           }           return fibThree;   } /*下面介绍一种时间复杂度是O(logn)的方法:对于斐波那契数列1,1,2,3,5,8,13…….有如下定义:F( n ) = F( n-1 ) + F( n-2 )F( 1 ) = 1F( 2 ) = 1矩阵形式：[ F( n+1 ) ,  F( n ) ] = [ F( n ) , F( n-1 ) ] * Q  其中 [ F( n+1 ) ,  F( n ) ]为行向量，Q = { [ 1, 1 ]; [ 1, 0 ] }为矩阵则 [ F( n+1 ) , F( n ) ]=[ 1 , 0 ] * Qn , */ struct Matrix    {                 long long m_00, m_01, m_10, m_11;   　　 Matrix ( long long m00 = 0,  long long m01 = 0,  long long m10 = 0,   long long m11 = 0 )           :m_00( m00 ), m_01( m01 ), m_10( m10 ), m_11( m11 )            {           }    }; Matrix MatrixMultiply (  const Matrix & m1, const Matrix & m2    ）    {  　　long long m00 = m1.m_00 * m2.m_00 + m1.m_01 * m2.m_10;　　long long m01 = m1.m_00 * m2.m_01 + m1.m_01 * m2.m_11; 　　long long m10 = m1.m_10 * m2.m_00 + m1.m_11 * m2.m_10    　　long long m11 = m1.m_10 * m2.m_01 + m1.m_11 * m2.m_11;       return Matrix ( m00,  m01,  m10, m11 );    }Matrix MatrixPower( unsigned int n )    {           assert(n > 0);           Matrix m;           if( n == 1)           {                 m = Matrix(1, 1, 1, 0);           }          else if(n % 2 == 0)           {                 m = MatrixPower( n / 2 );                 m = MatrixMultiply( matrix, matrix );           }          else if( n % 2 == 1 )           {                 m = MatrixPower( (n - 1) / 2 );                 m = MatrixMultiply( m, m );                 m = MatrixMultiply( m, Matrix( 1, 1, 1, 0 ) );           }           return m;    }  long long Fibonacci( unsigned int n ){      int result[2] = { 0, 1 };      if( n < 2 )            return result[ n ];      Matrix Q = MatrixPower( n - 1 );　　//注意：按定义式应该用[ 1, 0 ]*Q, 或者等价于{ [ 1 , 0 ]; [ 0, 0 ] }*Q, 但是因为显然结果相同，所以略去这一步。      return Q.m_00;}