Leetcode-1137. N-th Tribonacci Number 第 N 个泰波那契数 (DP)

题目

泰波那契序列 Tn 定义如下：
$T_0=0$, $T_1=1$, $T_2=1$, 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 Tn 的值。
链接：https://leetcode.com/problems/n-th-tribonacci-number/

The Tribonacci sequence $T_n$ is defined as follows:

$T_0=0$, $T_1=1$, $T_2=1$, and $T_{n+3}=T_n+T_{n+1}+T_{n+2}$ for $n>=0$.

Given n, return the value of $T_n$.

Example:

Input: n = 4
Output: 4
Explanation:
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4

思路及代码

DP

• 同斐波那契Fibonacci

class Solution:
    def tribonacci(self, n: int) -> int:
        if n == 0:
            return 0
        if n == 1:
            return 1
        if n == 2:
            return 1
        t0 = 0
        t1 = 1
        t2 = 1
        for i in range(3, n+1):
            t = t0 + t1 + t2
            t0 = t1
            t1 = t2
            t2 = t
        return t

复杂度

T = $O(n)$
S = $O(1)$