Question

面试题 08.01. 三步问题

English Version

题目描述

三步问题。有个小孩正在上楼梯，楼梯有n阶台阶，小孩一次可以上1阶、2阶或3阶。实现一种方法，计算小孩有多少种上楼梯的方式。结果可能很大，你需要对结果模1000000007。

示例1:

 输入：n = 3 
 输出：4
 说明: 有四种走法

示例2:

 输入：n = 5
 输出：13

提示:

1. n范围在[1, 1000000]之间

解法

方法一：递推

我们定义 $f[i]$ 表示上第 $i$ 阶台阶的方法数，初始时 $f[1]=1$, $f[2]=2$, $f[3]=4$。答案为 $f[n]$。

递推公式为 $f[i] = f[i-1] + f[i-2] + f[i-3]$。

由于 $f[i]$ 只与 $f[i-1]$, $f[i-2]$, $f[i-3]$ 有关，因此我们可以使用三个变量 $a$, $b$, $c$ 来存储 $f[i-1]$, $f[i-2]$, $f[i-3]$ 的值，使得空间复杂度降低到 $O(1)$。

时间复杂度 $O(n)$，其中 $n$ 为给定的整数。空间复杂度 $O(1)$。

class Solution:
  def waysToStep(self, n: int) -> int:
    a, b, c = 1, 2, 4
    mod = 10**9 + 7
    for _ in range(n - 1):
      a, b, c = b, c, (a + b + c) % mod
    return a

class Solution {
  public int waysToStep(int n) {
    final int mod = (int) 1e9 + 7;
    int a = 1, b = 2, c = 4;
    for (int i = 1; i < n; ++i) {
      int t = a;
      a = b;
      b = c;
      c = (((a + b) % mod) + t) % mod;
    }
    return a;
  }
}

class Solution {
public:
  int waysToStep(int n) {
    const int mod = 1e9 + 7;
    int a = 1, b = 2, c = 4;
    for (int i = 1; i < n; ++i) {
      int t = a;
      a = b;
      b = c;
      c = (((a + b) % mod) + t) % mod;
    }
    return a;
  }
};

func waysToStep(n int) int {
  const mod int = 1e9 + 7
  a, b, c := 1, 2, 4
  for i := 1; i < n; i++ {
    a, b, c = b, c, (a+b+c)%mod
  }
  return a
}

impl Solution {
  pub fn ways_to_step(n: i32) -> i32 {
    let (mut a, mut b, mut c) = (1, 2, 4);
    let m = 1000000007;
    for _ in 1..n {
      let t = a;
      a = b;
      b = c;
      c = (((a + b) % m) + t) % m;
    }
    a
  }
}

/**
 * @param {number} n
 * @return {number}
 */
var waysToStep = function (n) {
  let [a, b, c] = [1, 2, 4];
  const mod = 1e9 + 7;
  for (let i = 1; i < n; ++i) {
    [a, b, c] = [b, c, (a + b + c) % mod];
  }
  return a;
};

int waysToStep(int n) {
  const int mod = 1e9 + 7;
  int a = 1, b = 2, c = 4;
  for (int i = 1; i < n; ++i) {
    int t = a;
    a = b;
    b = c;
    c = (((a + b) % mod) + t) % mod;
  }
  return a;
}

方法二：矩阵快速幂加速递推

我们设 $F(n)$ 表示一个 $1 \times 3$ 的矩阵 $\begin{bmatrix} F_{n - 1} & F_{n - 2} & F_{n - 3} \end{bmatrix}$，其中 $F_{n - 1}$, $F_{n - 2}$ 和 $F_{n - 3}$ 分别表示上第 $n - 1$ 阶、第 $n - 2$ 阶和第 $n - 3$ 阶台阶的方法数。

我们希望根据 $F(n-1) = \begin{bmatrix} F_{n - 2} & F_{n - 3} & F_{n - 4} \end{bmatrix}$ 推出 $F(n)$。也即是说，我们需要一个矩阵 $base$，使得 $F(n - 1) \times base = F(n)$，即：

$$ \begin{bmatrix} F_{n - 2} & T_{n - 3} & T_{n - 4} \end{bmatrix} \times base = \begin{bmatrix} F_{n - 1} & F_{n - 2} & F_{n - 3} \end{bmatrix} $$

由于 $F_n = F_{n - 1} + F_{n - 2} + F_{n - 3}$，所以矩阵 $base$ 为：

$$ \begin{bmatrix} 1 & 1 & 0 \ 1 & 0 & 1 \ 1 & 0 & 0 \end{bmatrix} $$

我们定义初始矩阵 $res = \begin{bmatrix} 1 & 1 & 0 \end{bmatrix}$，那么 $F_n$ 等于 $res$ 乘以 $base^{n - 4}$ 的结果矩阵中所有元素之和。使用矩阵快速幂求解即可。

时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。

class Solution:
  def waysToStep(self, n: int) -> int:
    mod = 10**9 + 7

    def mul(a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
      m, n = len(a), len(b[0])
      c = [[0] * n for _ in range(m)]
      for i in range(m):
        for j in range(n):
          for k in range(len(a[0])):
            c[i][j] = (c[i][j] + a[i][k] * b[k][j] % mod) % mod
      return c

    def pow(a: List[List[int]], n: int) -> List[List[int]]:
      res = [[4, 2, 1]]
      while n:
        if n & 1:
          res = mul(res, a)
        n >>= 1
        a = mul(a, a)
      return res

    if n < 4:
      return 2 ** (n - 1)
    a = [[1, 1, 0], [1, 0, 1], [1, 0, 0]]
    return sum(pow(a, n - 4)[0]) % mod

class Solution {
  private final int mod = (int) 1e9 + 7;

  public int waysToStep(int n) {
    if (n < 4) {
      return (int) Math.pow(2, n - 1);
    }
    long[][] a = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
    long[][] res = pow(a, n - 4);
    long ans = 0;
    for (long x : res[0]) {
      ans = (ans + x) % mod;
    }
    return (int) ans;
  }

  private long[][] mul(long[][] a, long[][] b) {
    int m = a.length, n = b[0].length;
    long[][] c = new long[m][n];
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        for (int k = 0; k < b.length; ++k) {
          c[i][j] = (c[i][j] + a[i][k] * b[k][j] % mod) % mod;
        }
      }
    }
    return c;
  }

  private long[][] pow(long[][] a, int n) {
    long[][] res = {{4, 2, 1}};
    while (n > 0) {
      if ((n & 1) == 1) {
        res = mul(res, a);
      }
      a = mul(a, a);
      n >>= 1;
    }
    return res;
  }
}

class Solution {
public:
  int waysToStep(int n) {
    if (n < 4) {
      return pow(2, n - 1);
    }
    vector<vector<ll>> a = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
    vector<vector<ll>> res = qpow(a, n - 4);
    ll ans = 0;
    for (ll x : res[0]) {
      ans = (ans + x) % mod;
    }
    return ans;
  }

private:
  using ll = long long;
  const int mod = 1e9 + 7;
  vector<vector<ll>> mul(vector<vector<ll>>& a, vector<vector<ll>>& b) {
    int m = a.size(), n = b[0].size();
    vector<vector<ll>> c(m, vector<ll>(n));
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        for (int k = 0; k < b.size(); ++k) {
          c[i][j] = (c[i][j] + a[i][k] * b[k][j] % mod) % mod;
        }
      }
    }
    return c;
  }

  vector<vector<ll>> qpow(vector<vector<ll>>& a, int n) {
    vector<vector<ll>> res = {{4, 2, 1}};
    while (n) {
      if (n & 1) {
        res = mul(res, a);
      }
      a = mul(a, a);
      n >>= 1;
    }
    return res;
  }
};

const mod = 1e9 + 7

func waysToStep(n int) (ans int) {
  if n < 4 {
    return int(math.Pow(2, float64(n-1)))
  }
  a := [][]int{{1, 1, 0}, {1, 0, 1}, {1, 0, 0}}
  res := pow(a, n-4)
  for _, x := range res[0] {
    ans = (ans + x) % mod
  }
  return
}

func mul(a, b [][]int) [][]int {
  m, n := len(a), len(b[0])
  c := make([][]int, m)
  for i := range c {
    c[i] = make([]int, n)
  }
  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      for k := 0; k < len(b); k++ {
        c[i][j] = (c[i][j] + a[i][k]*b[k][j]%mod) % mod
      }
    }
  }
  return c
}

func pow(a [][]int, n int) [][]int {
  res := [][]int{{4, 2, 1}}
  for n > 0 {
    if n&1 == 1 {
      res = mul(res, a)
    }
    a = mul(a, a)
    n >>= 1
  }
  return res
}

/**
 * @param {number} n
 * @return {number}
 */

const mod = 1e9 + 7;

var waysToStep = function (n) {
  if (n < 4) {
    return Math.pow(2, n - 1);
  }
  const a = [
    [1, 1, 0],
    [1, 0, 1],
    [1, 0, 0],
  ];
  let ans = 0;
  const res = pow(a, n - 4);
  for (const x of res[0]) {
    ans = (ans + x) % mod;
  }
  return ans;
};

function mul(a, b) {
  const [m, n] = [a.length, b[0].length];
  const c = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; ++j) {
      for (let k = 0; k < b.length; ++k) {
        c[i][j] =
          (c[i][j] + Number((BigInt(a[i][k]) * BigInt(b[k][j])) % BigInt(mod))) % mod;
      }
    }
  }
  return c;
}

function pow(a, n) {
  let res = [[4, 2, 1]];
  while (n) {
    if (n & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}

方法三

import numpy as np


class Solution:
  def waysToStep(self, n: int) -> int:
    if n < 4:
      return 2 ** (n - 1)
    mod = 10**9 + 7
    factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
    res = np.mat([(4, 2, 1)], np.dtype("O"))
    n -= 4
    while n:
      if n & 1:
        res = res * factor % mod
      factor = factor * factor % mod
      n >>= 1
    return res.sum() % mod