From c911b6b65de08531115e19c465ae8ea519fe44af Mon Sep 17 00:00:00 2001
From: SambacFeng <56753082+SambacFeng@users.noreply.github.com>
Date: Fri, 17 Sep 2021 16:40:17 +0800
Subject: [PATCH 1/2] =?UTF-8?q?=E4=BF=AE=E6=AD=A3=E9=94=99=E5=88=AB?=
 =?UTF-8?q?=E5=AD=97?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/数组理论基础.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/problems/数组理论基础.md b/problems/数组理论基础.md
index 6d7b9f9a..146cd2de 100644
--- a/problems/数组理论基础.md
+++ b/problems/数组理论基础.md
@@ -88,7 +88,7 @@ int main() {
 
 **所以可以看出在C++中二维数组在地址空间上是连续的**。
 
-像Java是没有指针的，同时也不对程序员暴漏其元素的地址，寻址操作完全交给虚拟机。
+像Java是没有指针的，同时也不对程序员暴露其元素的地址，寻址操作完全交给虚拟机。
 
 所以看不到每个元素的地址情况，这里我以Java为例，也做一个实验。
 

From 3e069c66165d41e2eab3f3127de516dc4ef5cbc2 Mon Sep 17 00:00:00 2001
From: SambacFeng <56753082+SambacFeng@users.noreply.github.com>
Date: Fri, 17 Sep 2021 23:55:11 +0800
Subject: [PATCH 2/2] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200202.=E5=BF=AB?=
 =?UTF-8?q?=E4=B9=90=E6=95=B0?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0202.快乐数.md | 36 +++++++++++++++++++-----------------
 1 file changed, 19 insertions(+), 17 deletions(-)

diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md
index 710c824d..02db507a 100644
--- a/problems/0202.快乐数.md
+++ b/problems/0202.快乐数.md
@@ -159,26 +159,28 @@ func getSum(n int) int {
 javaScript:
 
 ```js
-function getN(n) {
-    if (n == 1 || n == 0) return n;
-    let res = 0;
-    while (n) {
-        res += (n % 10) * (n % 10);
-        n = parseInt(n / 10);
+var isHappy = function (n) {
+    let m = new Map()
+
+    const getSum = (num) => {
+        let sum = 0
+        while (n) {
+            sum += (n % 10) ** 2
+            n = Math.floor(n / 10)
+        }
+        return sum
+    }
+
+    while (true) {
+        // n出现过，证明已陷入无限循环
+        if (m.has(n)) return false
+        if (n === 1) return true
+        m.set(n, 1)
+        n = getSum(n)
     }
-    return res;
 }
 
-var isHappy = function(n) {
-    const sumSet = new Set();
-    while (n != 1 && !sumSet.has(n)) {
-        sumSet.add(n);
-        n = getN(n);
-    }
-    return n == 1;
-};
-
-// 使用环形链表的思想 说明出现闭环 退出循环
+// 方法二：使用环形链表的思想 说明出现闭环 退出循环
 var isHappy = function(n) {
     if (getN(n) == 1) return true;
     let a = getN(n), b = getN(getN(n));