From 9aeea4f16e50b176ac93163466f314b09003a692 Mon Sep 17 00:00:00 2001
From: xuxin <840198532@qq.com>
Date: Tue, 17 Mar 2026 13:50:22 +0800
Subject: [PATCH] =?UTF-8?q?=E8=BF=9B=E9=98=B6day08-=E9=80=92=E5=BD=92?=
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---
 .../src/com/inmind/digui02/Demo09.java        | 43 +++++++++++++++++++
 1 file changed, 43 insertions(+)
 create mode 100644 javaSE-day08/src/com/inmind/digui02/Demo09.java

diff --git a/javaSE-day08/src/com/inmind/digui02/Demo09.java b/javaSE-day08/src/com/inmind/digui02/Demo09.java
new file mode 100644
index 0000000..dc638ca
--- /dev/null
+++ b/javaSE-day08/src/com/inmind/digui02/Demo09.java
@@ -0,0 +1,43 @@
+package com.inmind.digui02;
+/*
+递归求阶乘的代码
+
+5! = 5*4*3*2*1
+4! = 4*3*2*1
+3! = 3*2*1
+2! = 2*1
+1! =1
+
+阶乘另一种计算方式:
+5! = 5*4!
+4! = 4*3!
+3! = 3*2!
+2! = 2*1!
+1! =1
+
+求阶乘的数学规律是  n! = n*(n-1)!
+
+核心思想:递归代码如何编写??
+    1.判断当前是否是结束条件,如果是就结束(return)
+    2.如果不是结束条件,那么按数学规律或者逻辑规律调用自己
+
+n=1就是结束条件
+求阶乘的数学规律是 n! = n*(n-1)!
+ */
+public class Demo09 {
+    public static void main(String[] args) {
+     //计算5的阶乘
+        System.out.println(getJieCheng(5));
+    }
+
+    //定义求n的阶乘的功能方法
+    public static int getJieCheng(int n) {
+        //1.判断当前是否是结束条件,如果是就结束(return)
+        if (n == 1) {
+            return 1;
+        }
+        //2.如果不是结束条件,那么按数学规律或者逻辑规律调用自己
+        return n*getJieCheng(n-1);
+    }
+
+}