zoukankan      html  css  js  c++  java

  • 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. }
  • 相关阅读:
    [LeetCode] 226. Invert Binary Tree
    [LeetCode] 101. Symmetric Tree
    [LeetCode] 100. Same Tree
    [LeetCode] 104. Maximum Depth of Binary Tree
    [LeetCode] 280. Wiggle Sort
    [LeetCode] 42. Trapping Rain Water
    [LeetCode] 190. Reverse Bits
    [LeetCode] 144. Binary Tree Preorder Traversal
    [Leetcode] 58. Length of Last Word
    [LeetCode] 16. 3Sum Closest
  • 原文地址：https://www.cnblogs.com/zhuzhenfeng/p/4664484.html
Copyright © 2011-2022 走看看