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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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<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>
|
||
|
||
<p>如果所有断言都通过，那么您的题解将被 <strong>通过</strong>。</p>
|
||
<p><strong>示例 1：</strong></p>
|
||
<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>
|
||
|
||
<p><strong>示例 2：</strong></p>
|
||
<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>
|
||
|
||
<p><strong>提示：</strong></p>
|
||
<ul>
|
||
<li><code>1 <= nums.length <= 3 * 104</code></li>
|
||
<li><code>-104 <= nums[i] <= 104</code></li>
|
||
<li><code>nums</code> 已按 <strong>升序</strong> 排列</li>
|
||
</ul>
|
||
<p><strong>代码实现</strong></p>
|
||
<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 == <span class="number">0</span>) { <span class="comment">//数组的长度是零就直接的退出</span></span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="type">int</span> <span class="variable">fast</span> <span class="operator">=</span> <span class="number">1</span>, slow = <span class="number">1</span>; <span class="comment">//定义两个快慢指针</span></span><br><span class="line"> <span class="keyword">while</span> (fast < n) {</span><br><span class="line"> <span class="comment">//如果后面一个数和前面的那个数相同 慢指针不动 快指针动</span></span><br><span class="line"> <span class="comment">//然后下一次判断的时候 慢指针就别快指针慢了一次 再将慢指针指向快指针</span></span><br><span class="line"> <span class="comment">//相同的数就后移了</span></span><br><span class="line"> <span class="keyword">if</span> (nums[fast] != nums[fast - <span class="number">1</span>]) {</span><br><span class="line"> nums[slow] = nums[fast];</span><br><span class="line"> ++slow;</span><br><span class="line"> }</span><br><span class="line"> ++fast;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> slow;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
|
||
|
||
<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>
|
||
<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>
|
||
<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>
|
||
<p><strong>说明:</strong></p>
|
||
<p>为什么返回数值是整数，但输出的答案是数组呢?</p>
|
||
<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>
|
||
<p>你可以想象内部操作如下:</p>
|
||
<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>
|
||
|
||
<p><strong>示例 1：</strong></p>
|
||
<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>
|
||
|
||
<p><strong>示例 2：</strong></p>
|
||
<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>
|
||
|
||
<p><strong>提示：</strong></p>
|
||
<ul>
|
||
<li><code>0 <= nums.length <= 100</code></li>
|
||
<li><code>0 <= nums[i] <= 50</code></li>
|
||
<li><code>0 <= val <= 100</code></li>
|
||
</ul>
|
||
<p><strong>代码实现</strong></p>
|
||
<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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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="card-widget card-recent-post"><div class="item-headline"><i class="fas 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