剑指offer笔记面试题10----斐波那契数列

题目：求斐波那契数列的第n项。写一个函数，输入n，求斐波那契数列的第n项。斐波那契数列的定义如下：f(0) = 0, f(1) = 1,f(n) = f(n - 1) + f(n - 2)。

测试用例：

• 功能测试（如输入3、5、10等）。
• 边界值测试（如输入0、1、2）。
• 性能测试（输入较大的数字，如40、50、100等）。

测试代码：

void test(int n, int expected){
	if(fibonacci_solution1(n) == expected)
		printf("test for %d in solution1 passed.\n", n);
	else
		printf("test for %d in solution1 failed.\n", n);
	if(fibonacci_solution2(n) == expected)
		printf("test for %d in solution2 passed.\n", n);
	else
		printf("test for %d in solution2 failed.\n", n);
	if(fibonacci_solution3(n) == expected)
		printf("test for %d in solution3 passed.\n", n);
	else
		printf("test for %d in solution3 failed.\n", n);
}

本题考点：

• 考查应聘者对递归、循环的理解及编码能力。
• 考查应聘者对时间复杂度的分析能力。
• 如果面试官采用的是青蛙跳台阶的问题，那么同时还在考查应聘者的数学建模能力。

代码实现：

#include <cstdio>
#include <cassert>

//=================方法一：递归==================
long long fibonacci_solution1(unsigned int n){
	if(n <= 0)
		return 0;
	if(n == 1)
		return 1;
	return fibonacci_solution1(n - 1) + fibonacci_solution1(n - 2);
} 

//=================方法二：循环==================
long long fibonacci_solution2(unsigned n){
	int result[2] = {0, 1};
	if(n < 2)
		return result[n];
	long long fibMinusOne = 1;
	long long fibMinusTwo = 0;
	long long fibN = 0;
	for(unsigned int i = 2; i <= n; ++i){
		fibN = fibMinusOne + fibMinusTwo;
		fibMinusTwo = fibMinusOne;
		fibMinusOne = fibN;
	}
	return fibN;
} 

//=================方法三：基于矩阵乘法============= 
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, m_01, m_10, 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;
}
int main(){
	test(0, 0);
	test(1, 1);
	test(2, 1);
	test(3, 2);
	test(4, 3);
	test(5, 5);
	test(6, 8);
	test(7, 13);
	test(8, 21);
	test(9, 34);
	test(10, 55);
	test(40, 102334155);
	test(100, 3736710778780434371);
	return 0;
}