diff --git a/README.md b/README.md
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--- a/README.md
+++ b/README.md
@@ -284,6 +284,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 400 |[Nth Digit](https://leetcode.com/problems/nth-digit)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/400.%20Nth%20Digit.md)|Medium|  |
 | 404 |[Sum of Left Leaves](https://leetcode.com/problems/sum-of-left-leaves)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/404.%20Sum%20of%20Left%20Leaves.md)|Easy|  |
 | 405 |[Convert a Number to Hexadecimal](https://leetcode.com/problems/convert-a-number-to-hexadecimal)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/405.%20Convert%20a%20Number%20to%20Hexadecimal.md)|Easy|  |
+| 406 |[Queue Reconstruction by Height](https://leetcode.com/problems/queue-reconstruction-by-height/)|[C++](solutions/406.%20Queue%20Reconstruction%20by%20Height.md)|Medium|  |
 | 409 |[Longest Palindrome](https://leetcode.com/problems/longest-palindrome)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/409.%20Longest%20Palindrome.md)|Easy|  |
 | 412 |[Fizz Buzz](https://leetcode.com/problems/fizz-buzz/)|[C++](solutions/412.%20Fizz%20Buzz.md)|Easy|  |
 | 414 |[Third Maximum Number](https://leetcode.com/problems/third-maximum-number)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/414.%20Third%20Maximum%20Number.md)|Easy|  |
diff --git a/solutions/406. Queue Reconstruction by Height.md b/solutions/406. Queue Reconstruction by Height.md
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+# [406. Queue Reconstruction by Height](https://leetcode.com/problems/queue-reconstruction-by-height/)
+
+# 思路
+
+有这样一个队列，队列中的每个元素是一个pair，分别为一个人的身高和前面身高不低于当前身高的人的个数，现在这个队列被随机打乱了，让我们恢复这个队列。
+
+## 思路一
+
+假设我们现在已有了一个满足题意的队列`q`，然后新来了一个人`p`且这个人的身高小于（或等于）该队列中的所有身高，那我们应该怎样将新来的这个人插入到原队列呢？很简单，应该将这个人插入到位置`p[1]`处，即`q.insert(q.begin()+p[1], p)`;
+
+所以我们可以从一个空队列开始不断插入所有人，为了满足上面的条件“这个人的身高小于（或等于）该队列中的所有身高”，我们应该先对所有人进行排序，身高高的排前面，身高一样的话，则第二个数小的排前面。然后新建一个空的数组，遍历之前排好序的数组，然后根据每个元素的第二个数字，将其插入到 res 数组中对应的位置。
+
+由于数组的插入是O(n)的，所以总的时间复杂度为O(n^2)
+
+## 思路二
+
+思路一开辟了外的数组，其实我们可以在元素组上进行插入。还是先排序，然后从前往后遍历，将当前元素插入到前面的合适位置。这里的插入需要我们手动将需要移动的元素往后移一步。
+
+复杂度同思路一，但是实际测试要快不少。
+
+
+# C++
+## 思路一
+``` C++
+class Solution {
+public:
+    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
+        sort(people.begin(), people.end(), [](const vector<int>&a, const vector<int>&b){
+            return a[0] != b[0] ? a[0] > b[0] : a[1] < b[1];
+        });
+        vector<vector<int>>res;
+        for(auto p: people)
+            res.insert(res.begin() + p[1], p);
+        
+        return res;
+    }
+};
+```
+
+
+## 思路二
+``` C++
+class Solution {
+public:
+    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
+        sort(people.begin(), people.end(), [](const vector<int>&a, const vector<int>&b){
+            return a[0] != b[0] ? a[0] > b[0] : a[1] < b[1];
+        });
+        
+        int h, k;
+        for(int i = 1; i < people.size(); i++){
+            h = people[i][0]; k = people[i][1];
+            for(int j = i; j > k; j--){
+                people[j][0] = people[j-1][0];
+                people[j][1] = people[j-1][1];
+            }
+            people[k][0] = h; people[k][1] = k;
+        }
+        
+        return people;
+    }
+};
+```
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