爬楼梯算法

假设你正在爬楼梯。需要 n 阶你才能到达楼顶。

每次你可以爬 1 或 2 个台阶。你有多少种不同的方法可以爬到楼顶呢？

规律：斐波那契数列（** 数值过大会溢出）

C语言实现（迭代法）

#include
int main() {
    int n;
    scanf("%d", &n);

    if (n <=2) {
        printf("%d", n);
    }

    int preOne = 1;
    int preTwo = 2;
    int count = 0;
    for (int i = 3; i <= n; ++i) {
        count = preOne + preTwo;
        preOne = preTwo;
        preTwo = count;
    }

     printf("%d", count);
}

C语言（递归法）

#include
int dieDai(int n);
int diGui(int n);
int main() {
    int n;
    scanf("%d", &n);

    int count = diGui(n);

    printf("%dn", count);
}
//迭代
int dieDai(int n) {

    if (n <=2) {
        return n;
    }
    int preOne = 1;
    int preTwo = 2;
    int count = 0;
    for (int i = 3; i <= n; ++i) {
        count = preOne + preTwo;
        preOne = preTwo;
        preTwo = count;
    }

    return count;
}


//递归
int diGui(int n) {
    if (n == 1) {
        return 1;
    }

    if (n == 2) {
        return 2;
    }

    return diGui(n - 1) + diGui(n - 2);
}

JAVA语言实现

import java.util.Scanner;

public class FeiBo {

    public static void main(String[] args) {
        //楼梯数量
        Scanner scanner = new Scanner(System.in);

        int n = scanner.nextInt();

        int count = dieDai(n);
        System.out.println(count);
    }

    private static int dieDai(int n) {

        if (n <= 2 ) {
            return n;
        }

        int preOne = 1;
        int preTwo = 2;
        int count = 0;
        for (int i = 3; i <= n; i++) {
            count = preOne + preTwo;
            preOne = preTwo;
            preTwo = count;
        }

        return count;
    }
}