diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md
index 7910fc50..6401824b 100644
--- a/problems/0093.复原IP地址.md
+++ b/problems/0093.复原IP地址.md
@@ -227,7 +227,7 @@ private:
 public:
     vector<string> restoreIpAddresses(string s) {
         result.clear();
-        if (s.size() > 12) return result; // 算是剪枝了
+        if (s.size() < 4 || s.size() > 12) return result; // 算是剪枝了
         backtracking(s, 0, 0);
         return result;
     }
diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md
index 7ecca773..188ad3cb 100644
--- a/problems/0106.从中序与后序遍历序列构造二叉树.md
+++ b/problems/0106.从中序与后序遍历序列构造二叉树.md
@@ -103,7 +103,7 @@ TreeNode* traversal (vector<int>& inorder, vector<int>& postorder) {
 中序数组相对比较好切，找到切割点（后序数组的最后一个元素）在中序数组的位置，然后切割，如下代码中我坚持左闭右开的原则：
 
 
-```C++
+```CPP
 // 找到中序遍历的切割点
 int delimiterIndex;
 for (delimiterIndex = 0; delimiterIndex < inorder.size(); delimiterIndex++) {
@@ -130,7 +130,7 @@ vector<int> rightInorder(inorder.begin() + delimiterIndex + 1, inorder.end() );
 
 代码如下：
 
-```
+```CPP
 // postorder 舍弃末尾元素，因为这个元素就是中间节点，已经用过了
 postorder.resize(postorder.size() - 1);
 
@@ -144,7 +144,7 @@ vector<int> rightPostorder(postorder.begin() + leftInorder.size(), postorder.end
 
 接下来可以递归了，代码如下：
 
-```
+```CPP
 root->left = traversal(leftInorder, leftPostorder);
 root->right = traversal(rightInorder, rightPostorder);
 ```
diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md
index 00707347..5d9e8295 100644
--- a/problems/0383.赎金信.md
+++ b/problems/0383.赎金信.md
@@ -84,6 +84,10 @@ class Solution {
 public:
     bool canConstruct(string ransomNote, string magazine) {
         int record[26] = {0};
+        //add
+        if (ransomNote.size() > magazine.size()) {
+            return false;
+        }
         for (int i = 0; i < magazine.length(); i++) {
             // 通过recode数据记录 magazine里各个字符出现次数
             record[magazine[i]-'a'] ++;