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Commit ae6f9cd

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Merge pull request MisterBooo#81 from xiaoshuai96/master
solved @xiaoshuai96 
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> 本文首发于公众号「图解面试算法」,是 [图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>) 系列文章之一。
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>
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> 个人博客:https://www.zhangxiaoshuai.fun
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**本题选自leetcode中第1137题,easy级别,目前通过率52.4%**
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### 题目描述:
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```txt
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泰波那契序列 Tn 定义如下:
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T0 = 0, T1 = 1, T2 = 1, 且在 n >= 0 的条件下 Tn+3 = Tn + Tn+1 + Tn+2
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给你整数 n,请返回第 n 个泰波那契数 Tn 的值。
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示例 1:
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输入:n = 4
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输出:4
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解释:
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T_3 =わ 0 +たす 1 +たす 1 =わ 2
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T_4 =わ 1 +たす 1 +たす 2 =わ 4
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示例 2:
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输入:n = 25
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输出:1389537
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提示:
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0 <= n <= 37
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答案保证是一个 32 位整数,即 answer <= 2^31 - 1。
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```
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### 题目分析:
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要是之前有接触过斐波那契数列的话,这道题是非常容易有解决思路的。我们有以下三种方法(正经方法两种,哈哈哈)来解决该问题:
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```
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1.递归(但是leetcode中是无法AC的,超出时间限制,但是还是会将代码展示出来)
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2.动态规划(这种题都是已知前面的来求得未知的,使用dp再合适不过)
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3.暴力(抖机灵,看一乐就可以啦)
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```
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### GIF动画演示:
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![](../Animation/1137-Tribonacci.gif)
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## 代码:
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### 递归版本:
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```java
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public int tribonacci(int n) {
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if (n == 0) {
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return 0;
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}
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if (n == 1 || n == 2) {
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return 1;
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}
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return tribonacci(n - 1) + tribonacci(n - 2) + tribonacci(n -3);
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}
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```
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### 动态规划
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```java
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int[] dp = new int[38];
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public int tribonacci(int n) {
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if (dp[n] != 0) {
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return dp[n];
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}
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if (n == 0) {
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return 0;
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} else if (n == 1 || n == 2) {
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return 1;
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} else {
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int res = tribonacci(n - 1) + tribonacci(n - 2) + tribonacci(n - 3);
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dp[n] = res;
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return res;
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}
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}
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```
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### 暴力法(十分暴力,哈哈哈哈......)
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```java
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public int tribonacci(int n) {
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int[] Ts = {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, 2082876103};
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return Ts[n];
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}
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```
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