P74 Sword Finger Offer Interview Question 10: Fibonacci Sequence
Write a function, enter n, and find the nth term of the Fibonacci sequence.
Recursive solution:
Advantages: concise code
Disadvantages: too many repeated calculations, code efficiency is not high
// C语言代码
// 迭代解法
long long Fibonacci_iteration(unsigned n)
{
if (n <= 0)
return 0;
if (n == 1)
return 1;
return Fibonacci_iteration(n - 1) + Fibonacci_iteration(n - 2);
}
Loop solution:
Advantages: bottom-up calculation , time complexity is O(n)
// C语言代码
// 循环解法
long long Fibonacci(unsigned n)
{
int result[2] = {
0,1 };
if (n < 2)
return result[n];
long long fibNMinusOne = 1;
long long fibNMinusTwo = 0;
long long fibN = 0;
for (unsigned int i = 2; i <= n; i++)
{
fibN = fibNMinusOne + fibNMinusTwo;
fibNMinusTwo = fibNMinusOne;
fibNMinusOne = fibN;
}
return fibN;
}
Matrix solution:
Converting the Fibonacci sequence to the power of the matrix, the time complexity is O(logn) but not practical enough
#include <cassert>
struct Matrix2By2
{
Matrix2By2
(
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)
{
}
long long m_00;
long long m_01;
long long m_10;
long long m_11;
};
Matrix2By2 MatrixMultiply
(
const Matrix2By2& matrix1,
const Matrix2By2& matrix2
)
{
return Matrix2By2(
matrix1.m_00 * matrix2.m_00 + matrix1.m_01 * matrix2.m_10,
matrix1.m_00 * matrix2.m_01 + matrix1.m_01 * matrix2.m_11,
matrix1.m_10 * matrix2.m_00 + matrix1.m_11 * matrix2.m_10,
matrix1.m_10 * matrix2.m_01 + matrix1.m_11 * matrix2.m_11);
}
Matrix2By2 MatrixPower(unsigned int n)
{
assert(n > 0);
Matrix2By2 matrix;
if(n == 1)
{
matrix = Matrix2By2(1, 1, 1, 0);
}
else if(n % 2 == 0)
{
matrix = MatrixPower(n / 2);
matrix = MatrixMultiply(matrix, matrix);
}
else if(n % 2 == 1)
{
matrix = MatrixPower((n - 1) / 2);
matrix = MatrixMultiply(matrix, matrix);
matrix = MatrixMultiply(matrix, Matrix2By2(1, 1, 1, 0));
}
return matrix;
}
long long Fibonacci_Solution3(unsigned int n)
{
int result[2] = {
0, 1};
if(n < 2)
return result[n];
Matrix2By2 PowerNMinus2 = MatrixPower(n - 1);
return PowerNMinus2.m_00;
}
Expansion: the problem of frog jumping
A frog can jump up to one or two steps at a time. Find the total number of jumping methods the frog jumps on an n-level step.
// 力扣接口
// 剑指 Offer 10- II. 青蛙跳台阶问题
int numWays(int n)
{
if(n<2)
return 1;
long long fibNMinusOne = 1;
long long fibNMinusTwo = 1;
long long fibN = 0;
for(unsigned int i = 2; i <= n; ++ i)
{
fibN = (fibNMinusOne + fibNMinusTwo) % 1000000007;
fibNMinusTwo = fibNMinusOne;
fibNMinusOne = fibN;
}
return fibN;
}
Expansion 2: The problem of frog jumping
In the problem of frog jumping from steps, if the condition is changed to: a frog can jump up to one or two steps at a time... it can also jump up to n steps , at this time the frog can jump to an n step How many jumping methods are there in total?
