1137.第 N 个泰波那契数
题目描述
原题
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:
提示:
0 <= n <= 37
- 答案保证是一个 32 位整数,即
answer <= 2^31 - 1
。
题解
| class Solution {
public int tribonacci(int n) {
if(n == 0) return 0;
if(n < 3) return 1;
int first = 0;
int second = 1;
int third = 1;
for(int i = 3;i <= n;i++){
third = first + second + third;
second = third - first - second;//原来的third
first = third - second - first;//原来的second
}
return third;
}
}
|