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feat: update lc problems
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‎solution/0100-0199/0136.Single Number/README.md

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<!-- 这里写题目描述 -->
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<p>给定一个<strong>非空</strong>整数数组，除了某个元素只出现一次以外，其余每个元素均出现两次。找出那个只出现了一次的元素。</p>
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<p>给你一个 <strong>非空</strong> 整数数组 <code>nums</code> ，除了某个元素只出现一次以外，其余每个元素均出现两次。找出那个只出现了一次的元素。</p>
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<p><strong>说明：</strong></p>
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<p>你必须设计并实现线性时间复杂度的算法来解决此问题，且该算法只使用常量额外空间。</p>
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<p>你的算法应该具有线性时间复杂度。 你可以不使用额外空间来实现吗？</p>
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<div class="original__bRMd">
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<div>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<p><strong class="example">示例 1 ：</strong></p>
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<pre><strong>输入:</strong> [2,2,1]
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<strong>输出:</strong> 1
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<pre>
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<strong>输入：</strong>nums = [2,2,1]
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<strong>输出：</strong>1
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</pre>
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<p><strong>示例&nbsp;2:</strong></p>
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<p><strong class="example">示例 2 ：</strong></p>
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<pre><strong>输入:</strong> [4,1,2,1,2]
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<strong>输出:</strong> 4</pre>
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<pre>
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<strong>输入：</strong>nums = [4,1,2,1,2]
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<strong>输出：</strong>4
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</pre>
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<p><strong class="example">示例 3 ：</strong></p>
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<pre>
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<strong>输入：</strong>nums = [1]
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<strong>输出：</strong>1
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
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<li><code>-3 * 10<sup>4</sup> &lt;= nums[i] &lt;= 3 * 10<sup>4</sup></code></li>
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<li>除了某个元素只出现一次以外，其余每个元素均出现两次。</li>
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</ul>
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</div>
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</div>
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## 解法
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‎solution/0400-0499/0409.Longest Palindrome/README.md

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<strong>输入:</strong>1
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</pre>
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<p>&nbsp;</p>
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<p><strong>示例 3：</strong></p>
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<pre>
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<strong>输入:</strong>s = "aaaaaccc"
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<strong>输入:</strong>7</pre>
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<p><strong>提示:</strong></p>
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‎solution/0700-0799/0791.Custom Sort String/README.md

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<!-- 这里写题目描述 -->
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<p>给定两个字符串 <code>order</code> 和 <code>s</code> 。<code>order</code> 的所有单词都是 <strong>唯一</strong> 的，并且以前按照一些自定义的顺序排序。</p>
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<p>给定两个字符串 <code>order</code> 和 <code>s</code> 。<code>order</code> 的所有字母都是 <strong>唯一</strong> 的，并且以前按照一些自定义的顺序排序。</p>
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<p>对 <code>s</code> 的字符进行置换，使其与排序的&nbsp;<code>order</code>&nbsp;相匹配。更具体地说，如果在&nbsp;<code>order</code>&nbsp;中的字符 <code>x</code> 出现字符 <code>y</code> 之前，那么在排列后的字符串中， <code>x</code>&nbsp;也应该出现在 <code>y</code> 之前。</p>
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<p>返回 <em>满足这个性质的 <code>s</code> 的任意排列&nbsp;</em>。</p>
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<p>返回 <em>满足这个性质的 <code>s</code> 的任意一种排列&nbsp;</em>。</p>
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<p>&nbsp;</p>
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‎solution/0800-0899/0864.Shortest Path to Get All Keys/README.md

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<ul>
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<li><font color="#c7254e"><font face="Menlo, Monaco, Consolas, Courier New, monospace"><span style="font-size:12.6px"><span style="background-color:#f9f2f4">'.'</span></span></font></font> 代表一个空房间</li>
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<li><font color="#c7254e"><font face="Menlo, Monaco, Consolas, Courier New, monospace"><span style="font-size:12.6px"><span style="background-color:#f9f2f4">'#'</span></span></font></font> 代表一堵</li>
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<li><font color="#c7254e"><font face="Menlo, Monaco, Consolas, Courier New, monospace"><span style="font-size:12.6px"><span style="background-color:#f9f2f4">'#'</span></span></font></font> 代表一堵墙</li>
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<li><font color="#c7254e"><font face="Menlo, Monaco, Consolas, Courier New, monospace"><span style="font-size:12.6px"><span style="background-color:#f9f2f4">'@'</span></span></font></font>&nbsp;是起点</li>
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<li>小写字母代表钥匙</li>
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<li>大写字母代表锁</li>

‎solution/0800-0899/0890.Find and Replace Pattern/README.md

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public:
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vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
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vector<string> ans;
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for (auto& word : words)
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if (match(word, pattern))
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ans.push_back(word);
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auto match = [](string& s, string& t) {
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int m1[128] = {0};
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int m2[128] = {0};
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for (int i = 0; i < s.size(); ++i) {
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if (m1[s[i]] != m2[t[i]]) return 0;
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m1[s[i]] = i + 1;
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m2[t[i]] = i + 1;
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}
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return 1;
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};
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for (auto& word : words) if (match(word, pattern)) ans.emplace_back(word);
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return ans;
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}
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bool match(string s, string t) {
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vector<int> m1(128);
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vector<int> m2(128);
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for (int i = 0; i < s.size(); ++i) {
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if (m1[s[i]] != m2[t[i]]) return 0;
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m1[s[i]] = i + 1;
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m2[t[i]] = i + 1;
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}
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return 1;
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}
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};
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```
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‎solution/0800-0899/0890.Find and Replace Pattern/README_EN.md

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public:
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vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
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vector<string> ans;
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for (auto& word : words)
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if (match(word, pattern))
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ans.push_back(word);
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auto match = [](string& s, string& t) {
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int m1[128] = {0};
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int m2[128] = {0};
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for (int i = 0; i < s.size(); ++i) {
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if (m1[s[i]] != m2[t[i]]) return 0;
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m1[s[i]] = i + 1;
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m2[t[i]] = i + 1;
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}
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return 1;
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};
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for (auto& word : words) if (match(word, pattern)) ans.emplace_back(word);
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return ans;
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}
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bool match(string s, string t) {
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vector<int> m1(128);
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vector<int> m2(128);
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for (int i = 0; i < s.size(); ++i) {
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if (m1[s[i]] != m2[t[i]]) return 0;
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m1[s[i]] = i + 1;
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m2[t[i]] = i + 1;
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}
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return 1;
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}
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};
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```
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‎solution/0800-0899/0890.Find and Replace Pattern/Solution.cpp

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public:
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vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
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vector<string> ans;
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for (auto& word : words)
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if (match(word, pattern))
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ans.push_back(word);
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auto match = [](string& s, string& t) {
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int m1[128] = {0};
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int m2[128] = {0};
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for (int i = 0; i < s.size(); ++i) {
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if (m1[s[i]] != m2[t[i]]) return 0;
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m1[s[i]] = i + 1;
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m2[t[i]] = i + 1;
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}
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return 1;
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};
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for (auto& word : words) if (match(word, pattern)) ans.emplace_back(word);
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return ans;
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}
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bool match(string s, string t) {
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vector<int> m1(128);
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vector<int> m2(128);
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for (int i = 0; i < s.size(); ++i) {
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if (m1[s[i]] != m2[t[i]]) return 0;
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m1[s[i]] = i + 1;
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m2[t[i]] = i + 1;
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}
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return 1;
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}
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};

‎solution/0800-0899/0891.Sum of Subsequence Widths/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：排序 + 枚举贡献**
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**方法一：排序 + 枚举元素计算贡献**
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题目求解的是数组 `nums` 中所有子序列中最大值与最小值差值之和，注意到“子序列”，并且涉及到“最大值”与“最小值”，我们考虑先对数组 `nums` 进行排序。
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\end{aligned}
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$$
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我们将上述式子展开，可以得到：
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我们将上式展开，可以得到：
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$$
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\begin{aligned}
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\end{aligned}
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$$
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即：
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$$
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\begin{aligned}
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\sum_{i=0}^{n-1} (nums[i] - nums[n-i-1]) \times 2^i
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\end{aligned}
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$$
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因此我们只需要对数组 `nums` 进行排序，然后计算上述的贡献即可。注意答案的取模操作。
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时间复杂度 $O(n\times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 `nums` 的长度。

‎solution/1400-1499/1491.Average Salary Excluding the Minimum and Maximum Salary/README.md

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sum += salary[i];
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ma = max(ma, salary[i]);
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mi = min(mi, salary[i]);
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}
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}
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return (sum - mi - ma) * 1.0 / (salarySize - 2);
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}
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```

‎solution/1400-1499/1491.Average Salary Excluding the Minimum and Maximum Salary/README_EN.md

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sum += salary[i];
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ma = max(ma, salary[i]);
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mi = min(mi, salary[i]);
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}
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}
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return (sum - mi - ma) * 1.0 / (salarySize - 2);
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}
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```

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