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）
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 | 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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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 | 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; } }; }; |