斐波那契数列的递归和动态规划解法

#include <iostream>
using namespace std;

int fibonacci(int n) {
    if (n == 0 || n == 1) {
        return n;
    }
    return fibonacci(n-1) + fibonacci(n-2);
}

int main() {
    int n = 10;
    cout << "斐波那契数列的第" << n << "项为：" << fibonacci(n) << endl;
    return 0;
}

#include <iostream>
#include <vector>

using namespace std;

// 使用vector实现动态规划版本的斐波那契数列
int fibonacci(int n) {
	if (n <= 1) {
		return n;
	}

	vector<int> dp(n + 1, 0);
	dp[0] = 0;
	dp[1] = 1;

	for (int i = 2; i <= n; i++) {
		dp[i] = dp[i - 1] + dp[i - 2];
	}

	return dp[n];
}

int main() {
	int n = 10;
	cout << "第 " << n << " 个斐波那契数列的值为：" << fibonacci(n) << endl;
	return 0;
}

def fib(n):
    #递归解法求解斐波那契数列，但是效率较低，存在重复计算
    if n <= 1:
        return n
    return fib(n-1) + fib(n-2)

def fib_dp(n):
    #动态规划求解斐波那契数列，没有重复计算，结果保存数组，效率提升
    if n <= 1:
        return n
    f = [0] * (n+1)
    f[0], f[1] = 0, 1
    for i in range(2, n+1):
        f[i] = f[i-1] + f[i-2]
    return f[n]