JZ-C-09

剑指offer第九题：斐波那契数列

//============================================================================
// Name        : JZ-C-09.cpp
// Author      : Laughing_Lz
// Version     :
// Copyright   : All Right Reserved
// Description : 斐波那契数列
//============================================================================

#include <iostream>
using namespace std;
/**
 *输入n，返回第n个斐波那契数
 */
double Fibonacci(int n) {
    double fi;
    if (n <= 0) {
        return -1; //错误
    } else if (n == 1 || n == 2) {
        fi = 1;
    } else {
        fi = Fibonacci(n - 1) + Fibonacci(n - 2);//递归的时间复杂度是n的指数递增，还可能导致栈溢出
    }
    return fi;
}
long long Fibonacci1(int n) {
    long long previous[] = { 1, 1 };//避免重复计算，利用数组存储第n-1和n-2个数。时间复杂度为0(n)
    if (n <= 0) {
        return -1; //错误
    } else {
        if (n == 1 || n == 2) {
            return previous[n-1];
        }
        for (int m = 2; m < n; m++) {
            long long pre = previous[1];
            previous[1] = previous[0] + previous[1];
            previous[0] = pre;
        }
        return previous[1];
    }
}
int main() {
    long long result = Fibonacci1(50);
    cout << result << endl;
    return 0;
    return 0;
}