泰波那契序列 Tn 定义如下：
T0 = 0, T1 = 1, T2 = 1, 且在 n >= 0 的条件下 Tn+3 = Tn + Tn+1 + Tn+2
给你整数 n，请返回第 n 个泰波那契数 Tn 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
class Solution:
def tribonacci(self, n: int) -> int:
if n == 0:
return 0
if n <= 2:
return 1
p = 0
q = r = 1
for i in range(3, n + 1):
s = p + q + r
p, q, r = q, r, s
return s
class Solution:
def tribonacci(self,n:int) ->int:
if n==0:
return 0
if n<=2:
return 1
p =0
q=r=1
for i in range(3,n+1):
s = p+q+r
p,q,r= q,r,s
return s
class Solution:
def tribonacci(self,n:int)->int:
if n==0:
return 0
if n<=2:
return 1
p =0
q=r=1
for i in range(3,n+1):
s =p +q+r
p,q,r = q,r,s
return s
class Solution:
def tribonacci(self,n:int) -> int:
if n ==0:
return 0
if n<=2:
return 1
p=0
q =r=1
for i in range(3,n+1):
s = p+q+r
p,q,r = q,r,s
return s
class Solution:
def tribonacci(self,n:int) -> int:
if n==0
if n<=2:
return 1
p=0
q =r=1
for i in range(3,n+1):
s =p+q+r
p,q,r = q,r,s
return s
class Solution:
def tribonacci(self,n:int) -> int:
if n==0:
return 0
if n<=2:
return 1
p =0
q=r=1
for i in range(3,n+1):
s = p +q +r
p,q,r = q,r,s
return s
