70, 爬楼梯

一、题目描述

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？

示例 1
输入: n = 2
输出: 2
解释: 有两种方法可以爬到楼顶。

1. 1 阶 + 1 阶
2. 2 阶

示例 2
输入: n = 3
输出: 3
解释: 有三种方法可以爬到楼顶。

1. 1 阶 + 1 阶 + 1 阶
2. 1 阶 + 2 阶
3. 2 阶 + 1 阶

提示

• 1 <= n <= 45

相关主题

• 记忆化搜索
• 数学
• 动态规划

二、题解

方法 1: 动态规划

pub fn climb_stairs(n: i32) -> i32 {
    let (mut p, mut q, mut r) = (0, 0, 1);

    for _ in 0..n {
        p = q;
        q = r;
        r = p + q;
    }

    r
}

public int climbStairs(int n) {
    int p = 0, q = 0, r = 1;

    for (int i = 0; i < n; i++) {
        p = q;
        q = r;
        r = p + q;
    }

    return r;
}

func climbStairs(n int) int {
    p, q, r := 0, 0, 1
    
    for range n {
        p = q
        q = r
        r = p + q
    }
    
    return r
}

方法 2: 矩阵快速幂

pub fn climb_stairs(mut n: i32) -> i32 {
    let mut m = vec![vec![1, 1], vec![1, 0]];

    let multiply = |a: &[Vec<i64>], b: &[Vec<i64>]| {
        let mut c = vec![vec![0; 2]; 2];
        for i in 0..2 {
            for j in 0..2 {
                c[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j];
            }
        }
        c
    };
    let mut pow = || {
        let mut ret = vec![vec![1, 0], vec![0, 1]];
        while n > 0 {
            if n & 1 == 1 {
                ret = multiply(&ret, &m);
            }
            n >>= 1;
            m = multiply(&m, &m);
        }
        ret
    };

    let res = pow();

    res[0][0] as i32
}

public int climbStairs(int n) {
    long[][] m = new long[][]{new long[]{1, 1}, new long[]{1, 0}};

    BiFunction<long[][], long[][], long[][]> multiply = (a, b) -> {
        long[][] c = new long[2][2];
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                c[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j];
            }
        }
        return c;
    };
    BiFunction<long[][], Integer, long[][]> pow = (_m, _n) -> {
        long[][] ret = new long[][]{new long[]{1, 0}, new long[]{0, 1}};
        while (_n > 0) {
            if ((_n & 1) == 1) {
                ret = multiply.apply(ret, _m);
            }
            _n >>= 1;
            _m = multiply.apply(_m, _m);
        }
        return ret;
    };

    long[][] res = pow.apply(m, n);

    return (int) res[0][0];
}

func climbStairs(n int) int {
    m := make([][]int, 2)
    m[0], m[1] = []int{1, 1}, []int{1, 0}
    
    multiply := func(a, b [][]int) [][]int {
        c := make([][]int, 2)
        c[0], c[1] = []int{0, 0}, []int{0, 0}
        
        for i := range 2 {
            for j := range 2 {
                c[i][j] = a[i][0]*b[0][j] + a[i][1]*b[1][j]
            }
        }
        
        return c
    }
    pow := func() [][]int {
        ret := make([][]int, 2)
        ret[0], ret[1] = []int{1, 0}, []int{0, 1}
        
        for n > 0 {
            if n&1 == 1 {
                ret = multiply(ret, m)
            }
            n >>= 1
            m = multiply(m, m)
        }
        
        return ret
    }
    
    res := pow()
    return res[0][0]
}

方法 3: 通项公式

pub fn climb_stairs(n: i32) -> i32 {
    let sqrt5 = 5_f64.sqrt();
    let fib_n = ((1.0 + sqrt5) / 2.0).powi(n + 1) - ((1.0 - sqrt5) / 2.0).powi(n + 1);

    return (fib_n / sqrt5).round() as i32;
}

public int climbStairs(int n) {
    double sqrt5 = Math.sqrt(5);
    double fibN = Math.pow((1 + sqrt5) / 2, n + 1) - Math.pow((1 - sqrt5) / 2, n + 1);

    return (int) Math.round(fibN / sqrt5);
}

func climbStairs(n int) int {
    sqrt5 := math.Sqrt(5)
    fibN := math.Pow((1+sqrt5)/2, float64(n+1)) - math.Pow((1-sqrt5)/2, float64(n+1))
    
    return int(math.Round(fibN / sqrt5))
}

方法 4: 排列组合

pub fn climb_stairs(n: i32) -> i32 {
    let n = n as i64;
    let calc = |mut a, b| {
        let mut ans = 1;
        for i in 1..=b {
            ans *= a;
            a -= 1;
            ans /= i;
        }
        ans
    };

    let mut res = 1;
    for i in 1..=n / 2 {
        res += calc(n - i, i);
    }

    res as i32
}

public int climbStairs(int n) {
    BiFunction<Integer, Integer, Long> calc = (a, b) -> {
        long ans = 1;
        for (int i = 1; i <= b; i++) {
            ans *= a;
            a--;
            ans /= i;
        }
        return ans;
    };

    long res = 1;
    for (int i = 1, times = n / 2; i <= times; i++) {
        res += calc.apply(n - i, i);
    }

    return (int) res;
}

func climbStairs(n int) int {
    calc := func(a, b int) int {
        ans := 1
        for i := 1; i <= b; i++ {
            ans *= a
            a--
            ans /= i
        }
        return ans
    }
    
    res := 1
    for i, times := 1, n/2; i <= times; i++ {
        res += calc(n-i, i)
    }
    
    return res
}