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面试题 10- II. 青蛙跳台阶问题

题目描述

一只青蛙一次可以跳上1级台阶，也可以跳上2级台阶。求该青蛙跳上一个 n 级的台阶总共有多少种跳法。

答案需要取模 1e9+7（1000000007），如计算初始结果为：1000000008，请返回 1。

示例 1：

输入：n = 2
输出：2

示例 2：

输入：n = 7
输出：21

示例 3：

输入：n = 0
输出：1

提示：

• 0 <= n <= 100

注意：本题与主站 70 题相同：https://leetcode.cn/problems/climbing-stairs/

 

解法

青蛙想上第 n 级台阶，可从第 n-1 级台阶跳一级上去，也可从第 n-2 级台阶跳两级上去，即：f(n) = f(n-1) + f(n-2)。递推求解即可。

Python3

class Solution:
    def numWays(self, n: int) -> int:
        a, b = 0, 1
        for _ in range(n):
            a, b = b, a + b
        return b % 1000000007

Java

class Solution {
    public int numWays(int n) {
        int a = 0, b = 1;
        for (int i = 0; i < n; ++i) {
            int c = (a + b) % 1000000007;
            a = b;
            b = c;
        }
        return b;
    }
}

C++

class Solution {
public:
    int numWays(int n) {
        int a = 0, b = 1;
        for (int i = 0; i < n; ++i) {
            int c = (a + b) % 1000000007;
            a = b;
            b = c;
        }
        return b;
    }
};

JavaScript

/**
 * @param {number} n
 * @return {number}
 */
var numWays = function (n) {
    let a = 0,
        b = 1;
    for (let i = 0; i < n; ++i) {
        const c = (a + b) % (1e9 + 7);
        a = b;
        b = c;
    }
    return b;
};

Go

func numWays(n int) int {
    a, b := 0, 1
    for i := 0; i < n; i++ {
        a, b = b, (a + b) % 1000000007
    }
    return b
}

TypeScript

function numWays(n: number): number {
    let a = 0;
    let b = 1;
    for (let i = 0; i < n; i++) {
        [a, b] = [b, (a + b) % 1000000007];
    }
    return b;
}

Rust

impl Solution {
    pub fn num_ways(n: i32) -> i32 {
        let mut tup = (0, 1);
        for _ in 0..n {
            tup = (tup.1, (tup.0 + tup.1) % 1000000007);
        }
        tup.1
    }
}

C#

public class Solution {
    public int NumWays(int n) {
        int a = 1, b = 1, tmp;
        for (int i = 0; i < n; i++) {
            tmp = a;
            a = b;
            b = (tmp + b) % 1000000007;
        }
        return a % 1000000007;
    }
}

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