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<p>​	本篇涉及：<strong>哈希表，双指针</strong></p>
<hr>
<h1 id="二-题目"><a href="#二-题目" class="headerlink" title="二.题目"></a><code>二.题目</code></h1><h2 id="1-无重复字符的最长子串"><a href="#1-无重复字符的最长子串" class="headerlink" title="1. 无重复字符的最长子串"></a>1. <a target="_blank" rel="noopener" href="https://leetcode.cn/problems/longest-substring-without-repeating-characters/">无重复字符的最长子串</a></h2><h3 id="题目描述："><a href="#题目描述：" class="headerlink" title="题目描述："></a>题目描述：</h3><p>​		给定一个字符串 <code>s</code> ，请你找出其中不含有重复字符的 <strong>最长子串</strong> 的长度。</p>
<h3 id="输入输出样例："><a href="#输入输出样例：" class="headerlink" title="输入输出样例："></a>输入输出样例：</h3><h4 id="示例-1"><a href="#示例-1" class="headerlink" title="示例 1:"></a>示例 1:</h4><blockquote>
<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">&gt;输入: s = &quot;abcabcbb&quot;</span><br><span class="line">&gt;输出: 3 </span><br><span class="line">&gt;解释: 因为无重复字符的最长子串是 &quot;abc&quot;，所以其长度为 3。</span><br></pre></td></tr></table></figure>
</blockquote>
<h4 id="示例-2"><a href="#示例-2" class="headerlink" title="示例 2:"></a>示例 2:</h4><blockquote>
<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: s = &quot;abcabcbb&quot;</span><br><span class="line">输出: 3 </span><br><span class="line">解释: 因为无重复字符的最长子串是 &quot;abc&quot;，所以其长度为 3。</span><br></pre></td></tr></table></figure>
</blockquote>
<h4 id="示例-3"><a href="#示例-3" class="headerlink" title="示例 3:"></a>示例 3:</h4><blockquote>
<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></pre></td><td class="code"><pre><span class="line">输入: s = &quot;pwwkew&quot;</span><br><span class="line">输出: 3</span><br><span class="line">解释: 因为无重复字符的最长子串是 &quot;wke&quot;，所以其长度为 3。</span><br><span class="line">  请注意，你的答案必须是 子串 的长度，&quot;pwke&quot; 是一个子序列，不是子串。</span><br></pre></td></tr></table></figure>
</blockquote>
<h4 id="提示"><a href="#提示" class="headerlink" title="提示:"></a>提示:</h4><ul>
<li><code>0 &lt;= s.length &lt;= 5 * 104</code></li>
<li><code>s</code> 由英文字母、数字、符号和空格组成</li>
</ul>
<h3 id="方法一：滑动窗口"><a href="#方法一：滑动窗口" class="headerlink" title="方法一：滑动窗口"></a>方法一：滑动窗口</h3><ul>
<li><p>我们使用两个指针表示字符串中的某个子串（或窗口）的左右边界</p>
</li>
<li><p>在每一步的操作中，我们会将左指针向右移动一格，表示 <strong>我们开始枚举下一个字符作为起始位置</strong></p>
</li>
<li><p>然后我们可以不断地向右移动右指针，但需要保证这两个指针对应的子串中没有重复的字符（使用哈希表）。</p>
</li>
<li><p>在移动结束后，这个子串就对应着 <strong>以左指针开始的，不包含重复字符的最长子串</strong>。我们记录下这个子串的长度；</p>
</li>
<li><p>将左指针的位置指向右指针的位置，右指针继续右移开始下一轮查找。</p>
</li>
</ul>
<p>在枚举结束后，我们找到的最长的子串的长度即为答案。</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></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * 我们还需要使用一种数据结构来判断 是否有重复的字符，常用的数据结构为哈希集合</span></span><br><span class="line"><span class="comment"> * 在左指针向右移动的时候，我们从哈希集合中移除一个字符，在右指针向右移动的时候，我们往哈希集合中添加一个字符。</span></span><br><span class="line"><span class="comment"> *</span></span><br><span class="line"><span class="comment"> * 复杂度分析:</span></span><br><span class="line"><span class="comment"> * 时间复杂度：O(N)，其中 N 是字符串的长度。左指针和右指针分别会遍历整个字符串一次。</span></span><br><span class="line"><span class="comment"> * 空间复杂度：O(∣Σ∣)，其中 Σ 表示字符集（即字符串中可以出现的字符），∣Σ∣ 表示字符集的大小。</span></span><br><span class="line"><span class="comment"> * 在本题中没有明确说明字符集，因此可以默认为所有 ASCII 码在 [0, 128)[0,128) 内的字符，</span></span><br><span class="line"><span class="comment"> * 即 ∣Σ∣=128。我们需要用到哈希集合来存储出现过的字符，而字符最多有 ∣Σ∣ 个，因此空间复杂度为 O(∣Σ∣)。</span></span><br><span class="line"><span class="comment"> *</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line">    <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">lengthOfLongestSubstring</span><span class="params">(String s)</span> &#123;</span><br><span class="line">        <span class="keyword">if</span> (s.length() == <span class="number">0</span>) <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">        HashMap&lt;Character, Integer&gt; map = <span class="keyword">new</span> <span class="title class_">HashMap</span>&lt;Character, Integer&gt;();</span><br><span class="line">        <span class="type">int</span> <span class="variable">max</span> <span class="operator">=</span> <span class="number">0</span>;<span class="comment">//最长子串长度</span></span><br><span class="line">        <span class="type">int</span> <span class="variable">left</span> <span class="operator">=</span> <span class="number">0</span>;<span class="comment">//滑动窗口左下标，i相当于滑动窗口右下标</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 &lt; s.length(); i++) &#123;</span><br><span class="line">            <span class="comment">//charAt() 方法用于返回指定索引处的字符。索引范围为从 0 到 length() - 1。</span></span><br><span class="line">            <span class="keyword">if</span> (map.containsKey(s.charAt(i))) &#123;</span><br><span class="line">                <span class="comment">//map.get():返回字符所对应的索引，当发现重复元素时，窗口左指针右移，直到不重复为止</span></span><br><span class="line">                <span class="comment">//map.get(&#x27;a&#x27;)=0,因为map中只有第一个a的下标，然后更新left指针到原来left的的下一位</span></span><br><span class="line">                left = Math.max(left, map.get(s.charAt(i)) + <span class="number">1</span>);</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">//更新map中a映射的下标</span></span><br><span class="line">            map.put(s.charAt(i), i);</span><br><span class="line">            <span class="comment">//比较两个参数的大小，返回较大的那个</span></span><br><span class="line">            max = Math.max(max, i - left + <span class="number">1</span>);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> max;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>


<h2 id="2-盛最多水的容器"><a href="#2-盛最多水的容器" class="headerlink" title="2. 盛最多水的容器"></a>2. <a target="_blank" rel="noopener" href="https://leetcode.cn/problems/container-with-most-water/">盛最多水的容器</a></h2><h3 id="题目描述：-1"><a href="#题目描述：-1" class="headerlink" title="题目描述："></a>题目描述：</h3><p>​		给定一个长度为 n 的整数数组 height 。有 n 条垂线，第 i 条线的两个端点是 (i, 0) 和 (i, height[i]) 。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。返回容器可以储存的最大水量。</p>
<p>说明：你不能倾斜容器。</p>
<h3 id="输入输出样例：-1"><a href="#输入输出样例：-1" class="headerlink" title="输入输出样例："></a>输入输出样例：</h3><h4 id="示例-1-1"><a href="#示例-1-1" class="headerlink" title="示例 1:"></a>示例 1:</h4><p><img src="https://testingcf.jsdelivr.net/gh/magic-H728/Store@master/blog-img/202207301721329.jpeg" alt="img"></p>
<blockquote>
<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">&gt;输入：[1,8,6,2,5,4,8,3,7]</span><br><span class="line">&gt;输出：49 </span><br><span class="line">&gt;解释：图中垂直线代表输入数组 [1,8,6,2,5,4,8,3,7]。在此情况下，容器能够容纳水（表示为蓝色部分）的最大值为 49。</span><br></pre></td></tr></table></figure>
</blockquote>
<h4 id="示例-2-1"><a href="#示例-2-1" class="headerlink" title="示例 2:"></a>示例 2:</h4><blockquote>
<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：height = [1,1]</span><br><span class="line">输出：1</span><br></pre></td></tr></table></figure>
</blockquote>
<h4 id="提示-1"><a href="#提示-1" class="headerlink" title="提示:"></a>提示:</h4><ul>
<li><code>n == height.length</code></li>
<li><code>2 &lt;= n &lt;= 105</code></li>
<li><code>0 &lt;= height[i] &lt;= 104</code></li>
</ul>
<h3 id="方法一：双指针"><a href="#方法一：双指针" class="headerlink" title="方法一：双指针"></a>方法一：双指针</h3><p>设两指针 i , j ，指向的水槽板高度分别为 h[i] , h[j] ，此状态下水槽面积为 S(i, j) 。由于可容纳水的高度由两板中的 <strong>短板</strong> 决定，因此可得如下 <strong>面积公式 ：</strong>    <code>S(i, j) = min(h[i], h[j]) × (j - i)</code></p>
<p><img src="https://testingcf.jsdelivr.net/gh/magic-H728/Store@master/blog-img/202207301724574.png" alt="Picture0.png"></p>
<p>在每个状态下，无论长板或短板向中间收窄一格，都会导致水槽 底边宽度 -1−1 变短：</p>
<ul>
<li>若向内 <strong>移动短板</strong> ，水槽的短板 min(h[i], h[j]) 可能变大，因此下个水槽的面积 <strong>可能增大</strong> 。</li>
<li>若向内 <strong>移动长板</strong> ，水槽的短板 min(h[i], h[j]) 不变或变小，因此下个水槽的面积 <strong>一定变小</strong> 。</li>
</ul>
<p>因此，初始化双指针分列水槽左右两端，循环每轮将短板向内移动一格，并更新面积最大值，直到两指针相遇时跳出；即可获得最大面积。</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></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * 1. 初始化： 双指针 i , j 分列水槽左右两端；</span></span><br><span class="line"><span class="comment"> * 2. 循环收窄： 直至双指针相遇时跳出；</span></span><br><span class="line"><span class="comment"> * 1. 选定两板高度中的短板，向中间收窄一格；</span></span><br><span class="line"><span class="comment"> * 2. 更新面积最大值 res ；</span></span><br><span class="line"><span class="comment"> * 3. 返回值： 返回面积最大值 res 即可；</span></span><br><span class="line"><span class="comment"> *</span></span><br><span class="line"><span class="comment"> * 复杂度分析：</span></span><br><span class="line"><span class="comment"> * 时间复杂度 O(N) ： 双指针遍历一次底边宽度 N。</span></span><br><span class="line"><span class="comment"> * 空间复杂度 O(1)： 变量 i , j , res 使用常数额外空间。</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"></span><br><span class="line"><span class="keyword">public</span> <span class="keyword">class</span> <span class="title class_">Problem11</span> &#123;</span><br><span class="line">    <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">maxArea</span><span class="params">(<span class="type">int</span>[] height)</span> &#123;</span><br><span class="line">        <span class="comment">// 初始化： 双指针 i , j 分列水槽左右两端；</span></span><br><span class="line">        <span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> <span class="number">0</span>, j = height.length - <span class="number">1</span>, res = <span class="number">0</span>;</span><br><span class="line">        <span class="comment">// 循环收窄： 直至双指针相遇时跳出；</span></span><br><span class="line">        <span class="keyword">while</span> (i &lt; j) &#123;</span><br><span class="line">            <span class="comment">// 选定两板高度中的短板，向中间收窄一格；</span></span><br><span class="line">            res = height[i] &lt; height[j] ?</span><br><span class="line">                    Math.max(res, (j - i) * height[i++]) :</span><br><span class="line">                    Math.max(res, (j - i) * height[j--]);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

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class="item-headline"><i></i><span></span></div><div class="item-content"></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"><ol class="toc"><li class="toc-item toc-level-1"><a class="toc-link" href="#%E4%B8%80-%E7%AE%80%E4%BB%8B"><span class="toc-text">一.简介</span></a></li><li class="toc-item toc-level-1"><a class="toc-link" href="#%E4%BA%8C-%E9%A2%98%E7%9B%AE"><span class="toc-text">二.题目</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#1-%E6%97%A0%E9%87%8D%E5%A4%8D%E5%AD%97%E7%AC%A6%E7%9A%84%E6%9C%80%E9%95%BF%E5%AD%90%E4%B8%B2"><span class="toc-text">1. 无重复字符的最长子串</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#%E9%A2%98%E7%9B%AE%E6%8F%8F%E8%BF%B0%EF%BC%9A"><span class="toc-text">题目描述：</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E8%BE%93%E5%85%A5%E8%BE%93%E5%87%BA%E6%A0%B7%E4%BE%8B%EF%BC%9A"><span class="toc-text">输入输出样例：</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#%E7%A4%BA%E4%BE%8B-1"><span class="toc-text">示例 1:</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#%E7%A4%BA%E4%BE%8B-2"><span class="toc-text">示例 2:</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#%E7%A4%BA%E4%BE%8B-3"><span class="toc-text">示例 3:</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#%E6%8F%90%E7%A4%BA"><span class="toc-text">提示:</span></a></li></ol></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E6%96%B9%E6%B3%95%E4%B8%80%EF%BC%9A%E6%BB%91%E5%8A%A8%E7%AA%97%E5%8F%A3"><span class="toc-text">方法一：滑动窗口</span></a></li></ol></li><li class="toc-item toc-level-2"><a class="toc-link" href="#2-%E7%9B%9B%E6%9C%80%E5%A4%9A%E6%B0%B4%E7%9A%84%E5%AE%B9%E5%99%A8"><span class="toc-text">2. 盛最多水的容器</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#%E9%A2%98%E7%9B%AE%E6%8F%8F%E8%BF%B0%EF%BC%9A-1"><span class="toc-text">题目描述：</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E8%BE%93%E5%85%A5%E8%BE%93%E5%87%BA%E6%A0%B7%E4%BE%8B%EF%BC%9A-1"><span class="toc-text">输入输出样例：</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#%E7%A4%BA%E4%BE%8B-1-1"><span class="toc-text">示例 1:</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#%E7%A4%BA%E4%BE%8B-2-1"><span class="toc-text">示例 2:</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#%E6%8F%90%E7%A4%BA-1"><span class="toc-text">提示:</span></a></li></ol></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E6%96%B9%E6%B3%95%E4%B8%80%EF%BC%9A%E5%8F%8C%E6%8C%87%E9%92%88"><span class="toc-text">方法一：双指针</span></a></li></ol></li></ol></li></ol></div></div><div class="card-widget card-recent-post"><div class="item-headline"><i class="fas fa-history"></i><span>最新文章</span></div><div class="aside-list"><div class="aside-list-item"><a class="thumbnail" href="/25737.html" title="数据结构OJ"><img src="https://testingcf.jsdelivr.net/gh/magic-H728/Store@master/blog-img/202301131413619.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="数据结构OJ"/></a><div class="content"><a class="title" href="/25737.html" title="数据结构OJ">数据结构OJ</a><time datetime="2023-01-13T00:00:00.000Z" title="发表于 2023-01-13 00:00:00">2023-01-13</time></div></div><div class="aside-list-item"><a class="thumbnail" href="/48665.html" title="Notion知识体系管理工具"><img src="https://testingcf.jsdelivr.net/gh/magic-H728/Store@master/blog-img/202207311705467.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="Notion知识体系管理工具"/></a><div class="content"><a class="title" href="/48665.html" 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