TribonacciNumber [翠波那契数]
# 介绍
泰波那契数列是斐波那契数列的一般化，其中每项都是前面三项的总和。其一般形式为 a(n) = a(n-1) + a(n-2) + a(n-3) 。
翠波那契序列：
0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852... 以此类推。
# 实现
一个简单的解决方案是递归。
// A simple recursive Javascript program to print first n Tribinocci numbers.
function printTribRec(n) {
if (n == 0 || n == 1 || n == 2) return 0;
if (n == 3) return 1;
return printTribRec(n - 1) + printTribRec(n - 2) + printTribRec(n - 3);
}
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使用动态规划：
Top-Down Dp Memoization：
function printTribRec(n, dp) {
if (n == 0 || n == 1 || n == 2) return 0;
if(dp[n] != -1) return dp[n];
if (n == 3) return 1;
return dp[n] = printTribRec(n - 1, dp) + printTribRec(n - 2, dp) + printTribRec(n - 3, dp);
}
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Bottom-Up DP Tabulation：
function printTrib(n) {
let dp = Array.from({length: n}, (_, i) => 0);
dp[0] = dp[1] = 0;
dp[2] = 1;
for (let i = 3; i < n; i++) dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3];
return dp;
}
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或者：
/**
* @function Tribonacci
* @description Tribonacci is the sum of previous three tribonacci numbers.
* @param {Integer} n - The input integer
* @return {Integer} tribonacci of n.
* @see [Tribonacci_Numbers](https://www.geeksforgeeks.org/tribonacci-numbers/)
*/
const tribonacci = (n) => {
// creating array to store previous tribonacci numbers
const dp = new Array(n + 1)
dp[0] = 0
dp[1] = 1
dp[2] = 1
for (let i = 3; i <= n; i++) dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3]
return dp[n]
}
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上述的时间复杂度是线性的，但它需要额外的空间。我们可以使用三个变量来跟踪前三个数字，以优化上述解决方案的空间。
// A space optimized based Javascript program to print first n Tribonacci numbers.
function printTrib(n) {
if (n < 1) return;
// Initialize first three numbers
let first = 0, second = 0, third = 1;
// Loop to add previous three numbers for each number starting from 3 and then assign first, second, third to second, third, and curr to third respectively
for (let i = 3; i < n; i++) {
let curr = first + second + third;
first = second;
second = third;
third = curr;
}
return third;
}
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更有效的解决方案：使用矩阵指数化。
// javascript Program to print first n tribonacci numbers
// Matrix Multiplication function for 3*3 matrix
function multiply(T , M) {
var a, b, c, d, e, f, g, h, i;
a = T[0][0] * M[0][0] + T[0][1] * M[1][0] + T[0][2] * M[2][0];
b = T[0][0] * M[0][1] + T[0][1] * M[1][1] + T[0][2] * M[2][1];
c = T[0][0] * M[0][2] + T[0][1] * M[1][2] + T[0][2] * M[2][2];
d = T[1][0] * M[0][0] + T[1][1] * M[1][0] + T[1][2] * M[2][0];
e = T[1][0] * M[0][1] + T[1][1] * M[1][1] + T[1][2] * M[2][1];
f = T[1][0] * M[0][2] + T[1][1] * M[1][2] + T[1][2] * M[2][2];
g = T[2][0] * M[0][0] + T[2][1] * M[1][0] + T[2][2] * M[2][0];
h = T[2][0] * M[0][1] + T[2][1] * M[1][1] + T[2][2] * M[2][1];
i = T[2][0] * M[0][2] + T[2][1] * M[1][2] + T[2][2] * M[2][2];
T[0][0] = a;
T[0][1] = b;
T[0][2] = c;
T[1][0] = d;
T[1][1] = e;
T[1][2] = f;
T[2][0] = g;
T[2][1] = h;
T[2][2] = i;
}
// Recursive function to raise the matrix T to the power n
function power(T , n) {
// base condition.
if (n == 0 || n == 1) return;
var M = [[ 1, 1, 1 ], [ 1, 0, 0 ], [ 0, 1, 0 ]];
// recursively call to square the matrix
power(T, parseInt(n / 2));
// calculating square of the matrix T
multiply(T, T);
// if n is odd multiply it one time with M
if (n % 2 != 0) multiply(T, M);
}
function tribonacci(n) {
var T = [[ 1, 1, 1 ], [ 1, 0, 0 ], [ 0, 1, 0 ]];
// base condition
if (n == 0 || n == 1) return 0;
else power(T, n - 2);
// T[0][0] contains the tribonacci number so return it
return T[0][0];
}
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参考：Matrix Exponentiation - GeeksforGeeks (opens new window)。
# 参考
编辑 (opens new window)
上次更新: 2022/10/27, 20:28:55