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 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& inorder, vector& 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 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 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'] ++;