From db84840986aab5ce791230eba963084bf1d5ba4b Mon Sep 17 00:00:00 2001 From: GitHubQAQ <31883473+GitHubQAQ@users.noreply.github.com> Date: Tue, 19 Apr 2022 14:23:08 +0800 Subject: [PATCH 1/3] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E9=A2=84=E5=88=A4?= =?UTF-8?q?=E6=96=AD=E9=95=BF=E5=BA=A6=E7=9A=84=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 在magazine中字符存入数组前，预先判断ransomNote和magazine的相对长度，进行处理。 --- problems/0383.赎金信.md | 4 ++++ 1 file changed, 4 insertions(+) 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'] ++; From dac0b4a12e43670eb54eaa6da5fa3f07cb707424 Mon Sep 17 00:00:00 2001 From: GitHubQAQ <31883473+GitHubQAQ@users.noreply.github.com> Date: Fri, 29 Apr 2022 21:03:27 +0800 Subject: [PATCH 2/3] =?UTF-8?q?Update=200106.=E4=BB=8E=E4=B8=AD=E5=BA=8F?= =?UTF-8?q?=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86=E5=BA=8F=E5=88=97?= =?UTF-8?q?=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 优化代码高亮 --- .../0106.从中序与后序遍历序列构造二叉树.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) 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 496de431..4396bc76 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); ``` From 29f41716f971506a001e8eb369b3f62964b43e8e Mon Sep 17 00:00:00 2001 From: GitHubQAQ <31883473+GitHubQAQ@users.noreply.github.com> Date: Sat, 7 May 2022 10:12:58 +0800 Subject: [PATCH 3/3] =?UTF-8?q?Update=200093.=E5=A4=8D=E5=8E=9FIP=E5=9C=B0?= =?UTF-8?q?=E5=9D=80.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 1. 优化C++版本剪枝 当字符串长度小于4时，无法组成正确的IP地址，直接返回。 关于判断是否为数字的部分，是否有必要（题目明确说明s仅由数字组成） [从解题角度可以删去，保留的话增加容错，个人意见] --- problems/0093.复原IP地址.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) 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; }