剑指offer：斐波那契数列

题目：写一个函数，输入n求斐波那契数列的第n项。


暴力简单解法：
  1. long long Fibonacci(unsigned int n)
  2. {
  3.     if(n<=0)
  4.         return 0;
  5.     if(n==1)
  6.         return 1;
  8.     return Fibonacci(n-1)+Fabonacci(n-2);
  9. }
利用循环改进之后，这个的时间复杂度是o（n)
  1. // ====================方法2：循环====================
  2. long long Fibonacci_Solution2(unsigned n)
  3. {
  4.     int result[2] = {0, 1};
  5.     if(n < 2)
  6.         return result[n];
  8.     long long  fibNMinusOne = 1;
  9.     long long  fibNMinusTwo = 0;
  10.     long long  fibN = 0;
  11.     for(unsigned int i = 2; i <= n; ++ i)
  12.     {
  13.         fibN = fibNMinusOne + fibNMinusTwo;
  15.         fibNMinusTwo = fibNMinusOne;
  16.         fibNMinusOne = fibN;
  17.     }
  19.      return fibN;
  20. }
第三种方法：利用如下的一个公式


  1. // ====================方法3：基于矩阵乘法====================
  2. #include <cassert>
  4. struct Matrix2By2
  5. {
  6.     Matrix2By2
  7.     (
  8.         long long m00 = 0, 
  9.         long long m01 = 0, 
  10.         long long m10 = 0, 
  11.         long long m11 = 0
  12.     )
  13.     :m_00(m00), m_01(m01), m_10(m10), m_11(m11) 
  14.     {
  15.     }
  17.     long long m_00;
  18.     long long m_01;
  19.     long long m_10;
  20.     long long m_11;
  21. };
  23. Matrix2By2 MatrixMultiply
  24. (
  25.     const Matrix2By2& matrix1, 
  26.     const Matrix2By2& matrix2
  27. )
  28. {
  29.     return Matrix2By2(
  30.         matrix1.m_00 * matrix2.m_00 + matrix1.m_01 * matrix2.m_10,
  31.         matrix1.m_00 * matrix2.m_01 + matrix1.m_01 * matrix2.m_11,
  32.         matrix1.m_10 * matrix2.m_00 + matrix1.m_11 * matrix2.m_10,
  33.         matrix1.m_10 * matrix2.m_01 + matrix1.m_11 * matrix2.m_11);
  34. }
  36. Matrix2By2 MatrixPower(unsigned int n)
  37. {
  38.     assert(n > 0);
  40.     Matrix2By2 matrix;
  41.     if(n == 1)
  42.     {
  43.         matrix = Matrix2By2(1, 1, 1, 0);
  44.     }
  45.     else if(n % 2 == 0)
  46.     {
  47.         matrix = MatrixPower(n / 2);
  48.         matrix = MatrixMultiply(matrix, matrix);
  49.     }
  50.     else if(n % 2 == 1)
  51.     {
  52.         matrix = MatrixPower((n - 1) / 2);
  53.         matrix = MatrixMultiply(matrix, matrix);
  54.         matrix = MatrixMultiply(matrix, Matrix2By2(1, 1, 1, 0));
  55.     }
  57.     return matrix;
  58. }
  60. long long Fibonacci_Solution3(unsigned int n)
  61. {
  62.     int result[2] = {0, 1};
  63.     if(n < 2)
  64.         return result[n];
  66.     Matrix2By2 PowerNMinus2 = MatrixPower(n - 1);
  67.     return PowerNMinus2.m_00;
  68. }






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原文地址：https://www.cnblogs.com/zhuzhenfeng/p/4664484.html