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# [1137. 第 N 个泰波那契数](https://leetcode-cn.com/problems/n-th-tribonacci-number)

[English Version](/solution/1100-1199/1137.N-th%20Tribonacci%20Number/README_EN.md)

## 题目描述

<p>泰波那契序列&nbsp;T<sub>n</sub>&nbsp;定义如下：&nbsp;</p>

<p>T<sub>0</sub> = 0, T<sub>1</sub> = 1, T<sub>2</sub> = 1, 且在 n &gt;= 0&nbsp;的条件下 T<sub>n+3</sub> = T<sub>n</sub> + T<sub>n+1</sub> + T<sub>n+2</sub></p>

<p>给你整数&nbsp;<code>n</code>，请返回第 n 个泰波那契数&nbsp;T<sub>n </sub>的值。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre><strong>输入：</strong>n = 4
<strong>输出：</strong>4
<strong>解释：</strong>
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4
</pre>

<p><strong>示例 2：</strong></p>

<pre><strong>输入：</strong>n = 25
<strong>输出：</strong>1389537
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>0 &lt;= n &lt;= 37</code></li>
	<li>答案保证是一个 32 位整数，即&nbsp;<code>answer &lt;= 2^31 - 1</code>。</li>
</ul>

## 解法

### **Python3**

```python
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
```

### **Java**

```java
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;
    }
}
```

### **C++**

```cpp
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;
    }
};
```

### **Go**

```go
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
}
```

### **JavaScript**

```js
/**
 * @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;
};
```

### **...**

```

```

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提交于 2021-07-05 19:43 . feat: add solutions to lc problems: No.0509,1137

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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

JavaScript

/**
 * @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;
};

...

Java

1

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