From 495225e1127bb426ab13bb9922aa298ba3251788 Mon Sep 17 00:00:00 2001
From: Omooo <869759698@qq.com>
Date: Wed, 17 Jun 2020 07:33:45 +0800
Subject: [PATCH] =?UTF-8?q?Update=20=E6=95=B0=E7=BB=84=E7=9B=B8=E5=85=B3.m?=
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---
 blogs/Algorithm/剑指 Offer/数组相关.md | 32 ++++++++++++++++++++++++++
 1 file changed, 32 insertions(+)

diff --git a/blogs/Algorithm/剑指 Offer/数组相关.md b/blogs/Algorithm/剑指 Offer/数组相关.md
index 48a6279..dbd933d 100644
--- a/blogs/Algorithm/剑指 Offer/数组相关.md	
+++ b/blogs/Algorithm/剑指 Offer/数组相关.md	
@@ -2,6 +2,19 @@
 数组相关
 ---
 
+#### 目录
+
+1. [03. 数组中重复的数字](https://leetcode-cn.com/problems/shu-zu-zhong-zhong-fu-de-shu-zi-lcof/)
+2. [04. 二维数组中的查找](https://leetcode-cn.com/problems/er-wei-shu-zu-zhong-de-cha-zhao-lcof/)
+3. [11. 旋转数组的最小数字](https://leetcode-cn.com/problems/xuan-zhuan-shu-zu-de-zui-xiao-shu-zi-lcof/)
+4. [21. 调整数组顺序使奇数位于偶数前面](https://leetcode-cn.com/problems/diao-zheng-shu-zu-shun-xu-shi-qi-shu-wei-yu-ou-shu-qian-mian-lcof/)
+5. [39. 数组中出现次数超过一半的数字](https://leetcode-cn.com/problems/shu-zu-zhong-chu-xian-ci-shu-chao-guo-yi-ban-de-shu-zi-lcof/)
+6. [42. 连续子数组的最大和](https://leetcode-cn.com/problems/lian-xu-zi-shu-zu-de-zui-da-he-lcof/)
+7. [45. 把数组排成最小的数](https://leetcode-cn.com/problems/ba-shu-zu-pai-cheng-zui-xiao-de-shu-lcof/)
+8. [56 - I. 数组中数字出现的次数](https://leetcode-cn.com/problems/shu-zu-zhong-shu-zi-chu-xian-de-ci-shu-lcof/)
+9. [56 - II. 数组中数字出现的次数 II](https://leetcode-cn.com/problems/shu-zu-zhong-shu-zi-chu-xian-de-ci-shu-ii-lcof/)
+10. [66. 构建乘积数组](https://leetcode-cn.com/problems/gou-jian-cheng-ji-shu-zu-lcof/)
+
 [03. 数组中重复的数字](https://leetcode-cn.com/problems/shu-zu-zhong-zhong-fu-de-shu-zi-lcof/)
 
 ```java
@@ -330,3 +343,22 @@ class Solution {
 }
 ```
 
+[66. 构建乘积数组](https://leetcode-cn.com/problems/gou-jian-cheng-ji-shu-zu-lcof/)
+
+```java
+class Solution {
+    public int[] constructArr(int[] a) {
+        int[] result = new int[a.length];
+        for (int i = 0, cur = 1; i < a.length; i++) {
+            result[i] = cur;
+            cur = cur * a[i];
+        }
+        for (int i = a.length - 1, cur = 1; i >= 0; i--) {
+            result[i] = result[i] * cur;
+            cur = cur * a[i];
+        }
+        return result;
+    }
+}
+```
+