动态规划（基础版）-- 斐波那契类型 -- 第 N 个泰波那契数

dynamic-programming/src/leetcode/editor/cn/NThTribonacciNumber.java

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+//<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>
+//
+//<div><div>Related Topics</div><div><li>记忆化搜索</li><li>数学</li><li>动态规划</li></div></div><br><div><li>👍 279</li><li>👎 0</li></div>
+package leetcode.editor.cn;
+
+// 1137:第 N 个泰波那契数
+public class NThTribonacciNumber {
+    public static void main(String[] args) {
+        Solution solution = new NThTribonacciNumber().new Solution();
+        // TO TEST
+    }
+
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int tribonacci(int n) {
+            if (n == 0) {
+                return 0;
+            } else if (n < 3) {
+                return 1;
+            }
+            int[] arr = new int[n + 1];
+            arr[1] = 1;
+            arr[2] = 1;
+            for (int i = 3; i <= n; i++) {
+                arr[i] = arr[i - 1] + arr[i - 2] + arr[i - 3];
+            }
+            return arr[n];
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

dynamic-programming/src/leetcode/editor/cn/doc/content/NThTribonacciNumber.md

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+<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>
+
+<div><div>Related Topics</div><div><li>记忆化搜索</li><li>数学</li><li>动态规划</li></div></div><br><div><li>👍 279</li><li>👎 0</li></div>