Question

1137. 第 N 个泰波那契数

English Version

题目描述

泰波那契序列 T_n 定义如下： 

T₀ = 0, T₁ = 1, T₂ = 1, 且在 n >= 0 的条件下 T_n+3 = T_n + T_n+1 + T_n+2

给你整数 n，请返回第 n 个泰波那契数 T_n 的值。

 

示例 1：

输入：n = 4
输出：4
解释：
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4

示例 2：

输入：n = 25
输出：1389537

 

提示：

• 0 <= n <= 37
• 答案保证是一个 32 位整数，即 answer <= 2^31 - 1。

解法

方法一：动态规划

根据题目中给出的递推式，我们可以使用动态规划求解。

我们定义三个变量 $a$, $b$, $c$，分别表示 $T_{n-3}$, $T_{n-2}$, $T_{n-1}$，初始值分别为 $0$, $1$, $1$。

然后从 $n$ 减小到 $0$，每次更新 $a$, $b$, $c$ 的值，直到 $n$ 为 $0$ 时，答案即为 $a$。

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

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

class Solution {
  public int tribonacci(int n) {
    int a = 0, b = 1, c = 1;
    while (n-- > 0) {
      int d = a + b + c;
      a = b;
      b = c;
      c = d;
    }
    return a;
  }
}

class Solution {
public:
  int tribonacci(int n) {
    long long a = 0, b = 1, c = 1;
    while (n--) {
      long long d = a + b + c;
      a = b;
      b = c;
      c = d;
    }
    return (int) a;
  }
};

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

function tribonacci(n: number): number {
  if (n === 0) {
    return 0;
  }
  if (n < 3) {
    return 1;
  }
  const a = [
    [1, 1, 0],
    [1, 0, 1],
    [1, 0, 0],
  ];
  return pow(a, n - 3)[0].reduce((a, b) => a + b);
}

function mul(a: number[][], b: number[][]): number[][] {
  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] += a[i][k] * b[k][j];
      }
    }
  }
  return c;
}

function pow(a: number[][], n: number): number[][] {
  let res = [[1, 1, 0]];
  while (n) {
    if (n & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}

/**
 * @param {number} n
 * @return {number}
 */
var tribonacci = function (n) {
  let a = 0;
  let b = 1;
  let c = 1;
  while (n--) {
    let d = a + b + c;
    a = b;
    b = c;
    c = d;
  }
  return a;
};

class Solution {
  /**
   * @param Integer $n
   * @return Integer
   */
  function tribonacci($n) {
    if ($n == 0) {
      return 0;
    } elseif ($n == 1 || $n == 2) {
      return 1;
    }
    $dp = [0, 1, 1];
    for ($i = 3; $i <= $n; $i++) {
      $dp[$i] = $dp[$i - 1] + $dp[$i - 2] + $dp[$i - 3];
    }
    return $dp[$n];
  }
}

方法二：矩阵快速幂加速递推

我们设 $Tib(n)$ 表示一个 $1 \times 3$ 的矩阵 $\begin{bmatrix} T_n & T_{n - 1} & T_{n - 2} \end{bmatrix}$，其中 $T_n$, $T_{n - 1}$ 和 $T_{n - 2}$ 分别表示第 $n$ 个、第 $n - 1$ 个和第 $n - 2$ 个泰波那契数。

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

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

由于 $T_n = T_{n - 1} + T_{n - 2} + T_{n - 3}$，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 $T_n$ 等于 $res$ 乘以 $base^{n - 3}$ 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 tribonacci(self, n: int) -> int:
    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]
      return c

    def pow(a: List[List[int]], n: int) -> List[List[int]]:
      res = [[1, 1, 0]]
      while n:
        if n & 1:
          res = mul(res, a)
        n >>= 1
        a = mul(a, a)
      return res

    if n == 0:
      return 0
    if n < 3:
      return 1
    a = [[1, 1, 0], [1, 0, 1], [1, 0, 0]]
    return sum(pow(a, n - 3)[0])

class Solution {
  public int tribonacci(int n) {
    if (n == 0) {
      return 0;
    }
    if (n < 3) {
      return 1;
    }
    int[][] a = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
    int[][] res = pow(a, n - 3);
    int ans = 0;
    for (int x : res[0]) {
      ans += x;
    }
    return ans;
  }

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

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

class Solution {
public:
  int tribonacci(int n) {
    if (n == 0) {
      return 0;
    }
    if (n < 3) {
      return 1;
    }
    vector<vector<ll>> a = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
    vector<vector<ll>> res = pow(a, n - 3);
    return accumulate(res[0].begin(), res[0].end(), 0);
  }

private:
  using ll = long long;
  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] += a[i][k] * b[k][j];
        }
      }
    }
    return c;
  }

  vector<vector<ll>> pow(vector<vector<ll>>& a, int n) {
    vector<vector<ll>> res = {{1, 1, 0}};
    while (n) {
      if (n & 1) {
        res = mul(res, a);
      }
      a = mul(a, a);
      n >>= 1;
    }
    return res;
  }
};

func tribonacci(n int) (ans int) {
  if n == 0 {
    return 0
  }
  if n < 3 {
    return 1
  }
  a := [][]int{{1, 1, 0}, {1, 0, 1}, {1, 0, 0}}
  res := pow(a, n-3)
  for _, x := range res[0] {
    ans += x
  }
  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] += a[i][k] * b[k][j]
      }
    }
  }
  return c
}

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

/**
 * @param {number} n
 * @return {number}
 */
var tribonacci = function (n) {
  if (n === 0) {
    return 0;
  }
  if (n < 3) {
    return 1;
  }
  const a = [
    [1, 1, 0],
    [1, 0, 1],
    [1, 0, 0],
  ];
  return pow(a, n - 3)[0].reduce((a, b) => a + b);
};

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] += a[i][k] * b[k][j];
      }
    }
  }
  return c;
}

function pow(a, n) {
  let res = [[1, 1, 0]];
  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 tribonacci(self, n: int) -> int:
    if n == 0:
      return 0
    if n < 3:
      return 1
    factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
    res = np.mat([(1, 1, 0)], np.dtype("O"))
    n -= 3
    while n:
      if n & 1:
        res *= factor
      factor *= factor
      n >>= 1
    return res.sum()