2023-09-22 21:57:28 +08:00
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2023-09-22 21:57:28 +08:00
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<meta property="og:description" content="刷题网站: 力扣 视频: AcWing 一.数组A.简单1.两数之和给定一个整数数组 nums 和一个整数目标值 target,请你在该数组中找出 和为目标值 target 的那 两个 整数,并返回它们的数组下标。你可以假设每种输入只会对应一个答案。但是,数组中同一个元素在答案里不能重复出现。你可以按任意顺序返回答案。 示例 1: 123输入:nums = [2,7,11,15], targe">
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2023-09-22 21:57:28 +08:00
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<h2 id="一-数组"><a href="#一-数组" class="headerlink" title="一.数组"></a>一.数组</h2><h3 id="A-简单"><a href="#A-简单" class="headerlink" title="A.简单"></a>A.简单</h3><h4 id="1-两数之和"><a href="#1-两数之和" class="headerlink" title="1.两数之和"></a>1.两数之和</h4><p>给定一个整数数组 <code>nums</code> 和一个整数目标值 <code>target</code>,请你在该数组中找出 <strong>和为目标值</strong> <em><code>target</code></em> 的那 <strong>两个</strong> 整数,并返回它们的数组下标。你可以假设每种输入只会对应一个答案。但是,数组中同一个元素在答案里不能重复出现。你可以按任意顺序返回答案。</p>
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<p><strong>示例 1:</strong></p>
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<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入:nums = [2,7,11,15], target = 9</span><br><span class="line">输出:[0,1]</span><br><span class="line">解释:因为 nums[0] + nums[1] == 9 ,返回 [0, 1] 。</span><br></pre></td></tr></table></figure>
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<p><strong>示例 2:</strong></p>
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<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入:nums = [3,2,4], target = 6</span><br><span class="line">输出:[1,2]</span><br></pre></td></tr></table></figure>
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<p><strong>示例 3:</strong></p>
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<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入:nums = [3,3], target = 6</span><br><span class="line">输出:[0,1]</span><br></pre></td></tr></table></figure>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>2 <= nums.length <= 104</code></li>
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<li><code>-109 <= nums[i] <= 109</code></li>
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<li><code>-109 <= target <= 109</code></li>
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<li><strong>只会存在一个有效答案</strong></li>
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</ul>
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<p><strong>代码实现</strong></p>
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<p>自己的代码 暴力枚举</p>
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<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span>[] twoSum(<span class="type">int</span>[] nums, <span class="type">int</span> target) {</span><br><span class="line"> <span class="type">int</span>[] result = <span class="keyword">new</span> <span class="title class_">int</span>[<span class="number">2</span>];</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> <span class="number">0</span>; i < nums.length; i++) {</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> <span class="variable">j</span> <span class="operator">=</span> <span class="number">0</span>; j < nums.length; j++) {</span><br><span class="line"> <span class="keyword">if</span> (nums[i] + nums[j] == target && i != j) {</span><br><span class="line"> result[<span class="number">0</span>] = i;</span><br><span class="line"> result[<span class="number">1</span>] = j;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> result;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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<h4 id="2-删除有序数组中的重复项"><a href="#2-删除有序数组中的重复项" class="headerlink" title="2.删除有序数组中的重复项"></a>2.删除有序数组中的重复项</h4><p>给你一个 <strong>升序排列</strong> 的数组 <code>nums</code> ,请你**<a target="_blank" rel="noopener" href="http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95"> 原地</a>** 删除重复出现的元素,使每个元素 <strong>只出现一次</strong> ,返回删除后数组的新长度。元素的 <strong>相对顺序</strong> 应该保持 <strong>一致</strong> 。然后返回 <code>nums</code> 中唯一元素的个数。</p>
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<p>考虑 <code>nums</code> 的唯一元素的数量为 <code>k</code> ,你需要做以下事情确保你的题解可以被通过:</p>
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<ul>
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<li>更改数组 <code>nums</code> ,使 <code>nums</code> 的前 <code>k</code> 个元素包含唯一元素,并按照它们最初在 <code>nums</code> 中出现的顺序排列。<code>nums</code> 的其余元素与 <code>nums</code> 的大小不重要。</li>
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<li>返回 <code>k</code> 。</li>
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</ul>
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<p><strong>判题标准:</strong></p>
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<p>系统会用下面的代码来测试你的题解:</p>
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<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line">int[] nums = [...]; // 输入数组</span><br><span class="line">int[] expectedNums = [...]; // 长度正确的期望答案</span><br><span class="line"></span><br><span class="line">int k = removeDuplicates(nums); // 调用</span><br><span class="line"></span><br><span class="line">assert k == expectedNums.length;</span><br><span class="line">for (int i = 0; i < k; i++) {</span><br><span class="line"> assert nums[i] == expectedNums[i];</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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<p>如果所有断言都通过,那么您的题解将被 <strong>通过</strong>。</p>
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<p><strong>示例 1:</strong></p>
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<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入:nums = [1,1,2]</span><br><span class="line">输出:2, nums = [1,2,_]</span><br><span class="line">解释:函数应该返回新的长度 2 ,并且原数组 nums 的前两个元素被修改为 1, 2 。不需要考虑数组中超出新长度后面的元素。</span><br></pre></td></tr></table></figure>
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<p><strong>示例 2:</strong></p>
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<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入:nums = [0,0,1,1,1,2,2,3,3,4]</span><br><span class="line">输出:5, nums = [0,1,2,3,4]</span><br><span class="line">解释:函数应该返回新的长度 5 , 并且原数组 nums 的前五个元素被修改为 0, 1, 2, 3, 4 。不需要考虑数组中超出新长度后面的元素。</span><br></pre></td></tr></table></figure>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 <= nums.length <= 3 * 104</code></li>
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<li><code>-104 <= nums[i] <= 104</code></li>
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<li><code>nums</code> 已按 <strong>升序</strong> 排列</li>
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</ul>
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<p><strong>代码实现</strong></p>
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<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">package</span> com.leetcode.array;</span><br><span class="line"></span><br><span class="line"><span class="keyword">import</span> java.util.Arrays;</span><br><span class="line"></span><br><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * <span class="doctag">@author</span> GongChangjiang</span></span><br><span class="line"><span class="comment"> * <span class="doctag">@version</span> 1.0</span></span><br><span class="line"><span class="comment"> * <span class="doctag">@Date</span> 2023/5/16</span></span><br><span class="line"><span class="comment"> * <span class="doctag">@Description</span> 删除有序数组中的重复项</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="keyword">public</span> <span class="keyword">class</span> <span class="title class_">RemoveDuplicates</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title function_">main</span><span class="params">(String[] args)</span> {</span><br><span class="line"> <span class="type">RemoveDuplicates</span> <span class="variable">removeDuplicates</span> <span class="operator">=</span> <span class="keyword">new</span> <span class="title class_">RemoveDuplicates</span>();</span><br><span class="line"> <span class="type">int</span>[] arr = {<span class="number">0</span>,<span class="number">0</span>,<span class="number">1</span>,<span class="number">1</span>,<span class="number">1</span>,<span class="number">2</span>,<span class="number">2</span>,<span class="number">3</span>,<span class="number">3</span>,<span class="number">4</span>};</span><br><span class="line"> System.out.println(removeDuplicates.removeDuplicates(arr));</span><br><span class="line"></span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> <span class="comment">/**</span></span><br><span class="line"><span class="comment"> * 数组去重</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">removeDuplicates</span><span class="params">(<span class="type">int</span>[] nums)</span> {</span><br><span class="line"> <span class="type">int</span> <span class="variable">n</span> <span class="operator">=</span> nums.length;<span class="comment">//获取数组长度</span></span><br><span class="line"> <span class="keyword">if</span> (n == <sp
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<h4 id="3-移除元素"><a href="#3-移除元素" class="headerlink" title="3.移除元素"></a>3.移除元素</h4><p>给你一个数组 <code>nums</code> 和一个值 <code>val</code>,你需要 <strong><a target="_blank" rel="noopener" href="https://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95">原地</a></strong> 移除所有数值等于 <code>val</code> 的元素,并返回移除后数组的新长度。</p>
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<p>不要使用额外的数组空间,你必须仅使用 <code>O(1)</code> 额外空间并 <strong><a target="_blank" rel="noopener" href="https://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95">原地 </a>修改输入数组</strong>。</p>
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<p>元素的顺序可以改变。你不需要考虑数组中超出新长度后面的元素。</p>
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<p><strong>说明:</strong></p>
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<p>为什么返回数值是整数,但输出的答案是数组呢?</p>
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<p>请注意,输入数组是以<strong>「引用」</strong>方式传递的,这意味着在函数里修改输入数组对于调用者是可见的。</p>
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<p>你可以想象内部操作如下:</p>
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<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><span class="line">// nums 是以“引用”方式传递的。也就是说,不对实参作任何拷贝</span><br><span class="line">int len = removeElement(nums, val);</span><br><span class="line"></span><br><span class="line">// 在函数里修改输入数组对于调用者是可见的。</span><br><span class="line">// 根据你的函数返回的长度, 它会打印出数组中 该长度范围内 的所有元素。</span><br><span class="line">for (int i = 0; i < len; i++) {</span><br><span class="line"> print(nums[i]);</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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<p><strong>示例 1:</strong></p>
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<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入:nums = [3,2,2,3], val = 3</span><br><span class="line">输出:2, nums = [2,2]</span><br><span class="line">解释:函数应该返回新的长度 2, 并且 nums 中的前两个元素均为 2。你不需要考虑数组中超出新长度后面的元素。例如,函数返回的新长度为 2 ,而 nums = [2,2,3,3] 或 nums = [2,2,0,0],也会被视作正确答案。</span><br></pre></td></tr></table></figure>
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<p><strong>示例 2:</strong></p>
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<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入:nums = [0,1,2,2,3,0,4,2], val = 2</span><br><span class="line">输出:5, nums = [0,1,4,0,3]</span><br><span class="line">解释:函数应该返回新的长度 5, 并且 nums 中的前五个元素为 0, 1, 3, 0, 4。注意这五个元素可为任意顺序。你不需要考虑数组中超出新长度后面的元素。</span><br></pre></td></tr></table></figure>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>0 <= nums.length <= 100</code></li>
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<li><code>0 <= nums[i] <= 50</code></li>
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<li><code>0 <= val <= 100</code></li>
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</ul>
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<p><strong>代码实现</strong></p>
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<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line"></span><br></pre></td></tr></table></figure>
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2024-06-14 22:00:25 +08:00
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</article><div class="tag_share"><div class="post-meta__tag-list"><a class="post-meta__tags" href="/tags/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/">数据结构与算法</a></div><div class="post_share"><div class="social-share" data-image="/img/6.png" data-sites="wechat,weibo,qq"></div><link rel="stylesheet" href="/cdn/css/share.min.css" media="print" onload="this.media='all'"><script src="/cdn/js/social-share.min.js" defer></script></div></div><div class="post-reward"><div class="reward-button"><i class="fas fa-qrcode"></i> 打赏</div><div class="reward-main"><ul class="reward-all"><li class="reward-item"><a href="/img/wechat.jpg" target="_blank"><img class="post-qr-code-img" src="/img/wechat.jpg" alt="微信"/></a><div class="post-qr-code-desc">微信</div></li><li class="reward-item"><a href="/img/alipay.jpg" target="_blank"><img class="post-qr-code-img" src="/img/alipay.jpg" alt="支付宝"/></a><div class="post-qr-code-desc">支付宝</div></li></ul></div></div><br/><div id="post-comment"><div class="comment-head"><div class="comment-headline"><i class="far fa-comment-alt fa-fw"></i><span> 评论</span></div></div><div class="comment-wrap"><div><div id="gitalk-container"></div></div></div></div></div><div class="aside-content" id="aside-content"><div class="card-widget card-info"><div class="is-center"><div class="avatar-img"><img src="/img/avatar.jpg" onerror="this.onerror=null;this.src='/img/loading.gif'" alt="avatar"/></div><div class="author-info__name">Jason</div><div class="author-info__description">Debug the World!</div></div><div class="card-info-data site-data is-center"><a href="/archives/"><div class="headline">文章</div><div class="length-num">60</div></a><a href="/tags/"><div class="headline">标签</div><div class="length-num">39</div></a><a href="/categories/"><div class="headline">分类</div><div class="length-num">10</div></a></div><a id="card-info-btn"><i class="fab fa-microsoft"></i><span>Ctrl + D 收藏</span></a><div class="card-info-social-icons is-center"><a class="social-icon" href="https://github.com/JasonsGong" target="_blank" title="Github"><i class="fab fa-github"></i></a><a class="social-icon" href="tencent://AddContact/?fromId=45&fromSubId=1&subcmd=all&uin=2602183349&website=www.oicqzone.com" target="_blank" title="QQ"><i class="fab fa-qq"></i></a><a class="social-icon" href="mailto:2602183349@qq.com" target="_blank" title="Email"><i class="fas fa-envelope-open-text"></i></a><a class="social-icon" href="https://github.com/JasonsGong?tab=repositories" target="_blank" title="代码仓库"><i class="fas fa-database"></i></a></div></div><div class="card-widget card-announcement"><div class="item-headline"><i class="fas fa-bullhorn fa-shake"></i><span>公告</span></div><div class="announcement_content">本网站是静态网站,更新页面资源请使用Ctrl+F5;若网站内文章对你有帮助,请使用Ctrl+D收藏该网站!</div></div><div class="sticky_layout"><div class="card-widget" id="card-toc"><div class="item-headline"><i class="fas fa-stream"></i><span>目录</span><span class="toc-percentage"></span></div><div class="toc-content is-expand"><ol class="toc"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E4%B8%80-%E6%95%B0%E7%BB%84"><span class="toc-text">一.数组</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#A-%E7%AE%80%E5%8D%95"><span class="toc-text">A.简单</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#1-%E4%B8%A4%E6%95%B0%E4%B9%8B%E5%92%8C"><span class="toc-text">1.两数之和</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#2-%E5%88%A0%E9%99%A4%E6%9C%89%E5%BA%8F%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9A%84%E9%87%8D%E5%A4%8D%E9%A1%B9"><span class="toc-text">2.删除有序数组中的重复项</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#3-%E7%A7%BB%E9%99%A4%E5%85%83%E7%B4%A0"><span class="toc-text">3.移除元素</span></a></li></ol></li></ol></li></ol></div></div><div class=
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2024-01-13 16:32:52 +08:00
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if (typeof Gitalk === 'function') initGitalk()
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else {
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2024-01-13 17:50:17 +08:00
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getCSS('/cdn/css/gitalk.min.css')
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getScript('/cdn/js/gitalk.min.js').then(initGitalk)
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2024-01-13 16:32:52 +08:00
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}
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}
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function commentCount(n){
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let isCommentCount = document.querySelector('#post-meta .gitalk-comment-count')
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if (isCommentCount) {
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isCommentCount.textContent= n
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}
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}
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if ('Gitalk' === 'Gitalk' || !true) {
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if (true) btf.loadComment(document.getElementById('gitalk-container'), loadGitalk)
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else loadGitalk()
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} else {
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function loadOtherComment () {
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loadGitalk()
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}
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}</script></div><script async data-pjax src="//busuanzi.ibruce.info/busuanzi/2.3/busuanzi.pure.mini.js"></script><div id="local-search"><div class="search-dialog"><nav class="search-nav"><span class="search-dialog-title">搜索</span><span id="loading-status"></span><button class="search-close-button"><i class="fas fa-times"></i></button></nav><div class="is-center" id="loading-database"><i class="fas fa-spinner fa-pulse"></i><span> 数据库加载中</span></div><div class="search-wrap"><div id="local-search-input"><div class="local-search-box"><input class="local-search-box--input" placeholder="搜索文章" type="text"/></div></div><br/><div class="no-result" id="local-search-results"></div><div id="local-search-stats-wrap"></div></div></div><div id="search-mask"></div><script src="/js/search/local-search.js"></script></div></div><!-- hexo injector body_end start --><script data-pjax>
|
2023-09-22 21:57:28 +08:00
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function butterfly_swiper_injector_config(){
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2024-01-13 22:42:28 +08:00
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var parent_div_git = document.getElementById('recent-posts');
|
2024-06-14 22:00:25 +08:00
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var item_html = '<div class="recent-post-item" style="height: auto;width: 100%"><div class="blog-slider swiper-container-fade swiper-container-horizontal" id="swiper_container"><div class="blog-slider__wrp swiper-wrapper" style="transition-duration: 0ms;"><div class="blog-slider__item swiper-slide" style="width: 750px; opacity: 1; transform: translate3d(0px, 0px, 0px); transition-duration: 0ms;"><a class="blog-slider__img" href="posts/19306.html" alt=""><img width="48" height="48" src="/img/1.png" alt="" onerror="this.src=https://unpkg.zhimg.com/akilar-candyassets/image/loading.gif; this.onerror = null;"/></a><div class="blog-slider__content"><span class="blog-slider__code">2023-04-21</span><a class="blog-slider__title" href="posts/19306.html" alt="">Docker容器化技术</a><div class="blog-slider__text">Docker</div><a class="blog-slider__button" href="posts/19306.html" alt="">详情 </a></div></div><div class="blog-slider__item swiper-slide" style="width: 750px; opacity: 1; transform: translate3d(0px, 0px, 0px); transition-duration: 0ms;"><a class="blog-slider__img" href="posts/47003.html" alt=""><img width="48" height="48" src="/img/5.png" alt="" onerror="this.src=https://unpkg.zhimg.com/akilar-candyassets/image/loading.gif; this.onerror = null;"/></a><div class="blog-slider__content"><span class="blog-slider__code">2023-03-10</span><a class="blog-slider__title" href="posts/47003.html" alt="">常用正则表达式大全</a><div class="blog-slider__text">正则表达式</div><a class="blog-slider__button" href="posts/47003.html" alt="">详情 </a></div></div><div class="blog-slider__item swiper-slide" style="width: 750px; opacity: 1; transform: translate3d(0px, 0px, 0px); transition-duration: 0ms;"><a class="blog-slider__img" href="posts/20683.html" alt=""><img width="48" height="48" src="/img/8.png" alt="" onerror="this.src=https://unpkg.zhimg.com/akilar-candyassets/image/loading.gif; this.onerror = null;"/></a><div class="blog-slider__content"><span class="blog-slider__code">2023-06-05</span><a class="blog-slider__title" href="posts/20683.html" alt="">Linux中开发环境的搭建</a><div class="blog-slider__text">环境搭建</div><a class="blog-slider__button" href="posts/20683.html" alt="">详情 </a></div></div><div class="blog-slider__item swiper-slide" style="width: 750px; opacity: 1; transform: translate3d(0px, 0px, 0px); transition-duration: 0ms;"><a class="blog-slider__img" href="posts/63333.html" alt=""><img width="48" height="48" src="/img/10.png" alt="" onerror="this.src=https://unpkg.zhimg.com/akilar-candyassets/image/loading.gif; this.onerror = null;"/></a><div class="blog-slider__content"><span class="blog-slider__code">2023-06-03</span><a class="blog-slider__title" href="posts/63333.html" alt="">开发环境的搭建</a><div class="blog-slider__text">环境搭建</div><a class="blog-slider__button" href="posts/63333.html" alt="">详情 </a></div></div></div><div class="blog-slider__pagination swiper-pagination-clickable swiper-pagination-bullets"></div></div></div>';
|
2024-01-13 22:42:28 +08:00
|
|
|
|
if (parent_div_git !== null && typeof parent_div_git !== 'undefined') {
|
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|
parent_div_git.insertAdjacentHTML("afterbegin",item_html)
|
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|
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|
}
|
2023-09-22 21:57:28 +08:00
|
|
|
|
}
|
|
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|
|
var elist = 'undefined'.split(',');
|
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|
var cpage = location.pathname;
|
2023-10-28 10:47:20 +08:00
|
|
|
|
var epage = 'all';
|
2023-09-22 21:57:28 +08:00
|
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|
var flag = 0;
|
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|
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|
for (var i=0;i<elist.length;i++){
|
|
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|
|
if (cpage.includes(elist[i])){
|
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|
flag++;
|
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|
}
|
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|
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|
}
|
|
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|
|
|
|
|
|
|
if ((epage ==='all')&&(flag == 0)){
|
|
|
|
|
butterfly_swiper_injector_config();
|
|
|
|
|
}
|
|
|
|
|
else if (epage === cpage){
|
|
|
|
|
butterfly_swiper_injector_config();
|
|
|
|
|
}
|
2024-01-13 22:10:58 +08:00
|
|
|
|
</script><script defer src="https://npm.elemecdn.com/hexo-butterfly-swiper/lib/swiper.min.js"></script><script defer data-pjax src="https://npm.elemecdn.com/hexo-butterfly-swiper/lib/swiper_init.js"></script><!-- hexo injector body_end end --></body></html>
|