时间复杂度为O( log n )的方法：
该算法使用矩阵乘法操作，使得算法时间复杂度为 O（logN）
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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;
} |
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class Solution {
public:
int Fibonacci(int n) {
int result[2]={0,1};
if(n<2)return result[n];
Matrix m;
return m.Power(n-1).a00;
}
class Matrix{
public:
long int a00;
long int a01;
long int a10;
long int a11;
Matrix (long int a,long int b,long int c,long int d){
a00=a;
a01=b;
a10=c;
a11=d;
}
Matrix (){
a00=1;a01=1;a10=1;a11=0;
}
Matrix operator * (Matrix & m2){
long int b00 = a00 * m2.a00 + a01 * m2.a10;
long int b01 = a00 * m2.a01 + a01 * m2.a11;
long int b10 = a10 * m2.a00 + a11 * m2.a10;
long int b11 = a10 * m2.a01 + a11 * m2.a11;
return Matrix(b00,b01,b10,b11);
}
Matrix Power( unsigned int n )
{
Matrix m;
if( n == 1)
{
m = Matrix(1, 1, 1, 0);
}
else if(n % 2 == 0)
{
m = Power( n / 2 );
m = m*m;
}
else if( n % 2 == 1 )
{
m = Power( (n - 1) / 2 );
m = m*m;
Matrix tmp;
m = m*tmp ;
}
return m;
}
};
}; |