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<p>这篇博客用于学习并记录在LeetCode刷题的过程及其相关题目解决思路。</p>
<a id="more"></a>
<h1 id="栈"><a href="#栈" class="headerlink" title="栈"></a>栈</h1><h2 id="20-有效的括号"><a href="#20-有效的括号" class="headerlink" title="20. 有效的括号"></a>20. <a href="https://leetcode-cn.com/problems/valid-parentheses/" rel="external nofollow" target="_blank">有效的括号</a></h2><p>判断字符串内括号顺序是否正确，<code>()[]</code>正确，<code>([)]</code>不正确，因为左方括号与右圆括号不合法。</p>
<p>解题思路1：</p>
<p>将左括号进栈，遍历到右括号就出栈，判断最后栈中是否还有剩余。</p>
<figure class="highlight python"><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></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">isValid</span><span class="params">(self, s: str)</span> -> bool:</span></span><br><span class="line"> <span class="keyword">if</span> len(s) % <span class="number">2</span> == <span class="number">1</span>:</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">False</span></span><br><span class="line"> d = {<span class="string">')'</span>:<span class="string">'('</span>, <span class="string">']'</span>:<span class="string">'['</span>, <span class="string">'}'</span>:<span class="string">'{'</span>}</span><br><span class="line"> tmp = [<span class="string">'t'</span>] <span class="comment"># 终结符，为了避免tmp为空时执行.pop方法出错</span></span><br><span class="line"> <span class="keyword">for</span> i <span class="keyword">in</span> s:</span><br><span class="line"> <span class="keyword">if</span> i <span class="keyword">in</span> d.keys():</span><br><span class="line"> <span class="keyword">if</span> d.get(i) != tmp.pop():</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">False</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> tmp.append(i)</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">False</span> <span class="keyword">if</span> len(tmp) != <span class="number">1</span> <span class="keyword">else</span> <span class="keyword">True</span></span><br></pre></td></tr></table></figure>
<p>解题思路2：</p>
<p>将<code>()</code>,<code>[]</code>,<code>{}</code>替换为空，如果最终字符串为空，则为真。</p>
<figure class="highlight python"><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></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">isValid</span><span class="params">(self, s: str)</span> -> bool:</span></span><br><span class="line"> <span class="keyword">if</span> len(s)%<span class="number">2</span>:</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">False</span></span><br><span class="line"> <span class="keyword">while</span> <span class="string">'()'</span> <span class="keyword">in</span> s <span class="keyword">or</span> <span class="string">'[]'</span> <span class="keyword">in</span> s <span class="keyword">or</span> <span class="string">'{}'</span> <span class="keyword">in</span> s:</span><br><span class="line"> s = s.replace(<span class="string">'[]'</span>,<span class="string">''</span>).replace(<span class="string">'()'</span>,<span class="string">''</span>).replace(<span class="string">'{}'</span>,<span class="string">''</span>)</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">False</span> <span class="keyword">if</span> len(s) <span class="keyword">else</span> <span class="keyword">True</span></span><br></pre></td></tr></table></figure>
<h2 id="42-接雨水"><a href="#42-接雨水" class="headerlink" title="42. 接雨水"></a>42. <a href="https://leetcode-cn.com/problems/trapping-rain-water/" rel="external nofollow" target="_blank">接雨水</a></h2><p>这道题挺有意思的，根据数组判断某个地形可容纳的积水。</p>
<p><img src="./leetcode/42.png" alt=""></p>
<figure class="highlight plain"><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: [0,1,0,2,1,0,1,3,2,1,2,1]</span><br><span class="line">输出: 6</span><br></pre></td></tr></table></figure>
<figure class="highlight python"><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></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">trap</span><span class="params">(self, height: List[int])</span> -> int:</span></span><br><span class="line"> <span class="string">'''</span></span><br><span class="line"><span class="string"> 思路：</span></span><br><span class="line"><span class="string"> 1. 先找到左侧制高点，将数组分为两部分，右边部分reversed一下</span></span><br><span class="line"><span class="string"> 2. 在两部分数组都执行“左低右高”式累加即可</span></span><br><span class="line"><span class="string"> '''</span></span><br><span class="line"> <span class="keyword">if</span> len(height) < <span class="number">3</span>: <span class="comment"># 数组至少有三个元素才能积水</span></span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span></span><br><span class="line"> max_idx = height.index(max(height))</span><br><span class="line"> h1 = height[:max_idx+<span class="number">1</span>]</span><br><span class="line"> h2 = list(reversed(height[max_idx:]))</span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">get_v</span><span class="params">(arr)</span>:</span></span><br><span class="line"> v = <span class="number">0</span></span><br><span class="line"> left = arr[<span class="number">0</span>]</span><br><span class="line"> cv = <span class="number">0</span> <span class="comment"># 某个区间的累计蓄水</span></span><br><span class="line"> <span class="keyword">for</span> h <span class="keyword">in</span> arr:</span><br><span class="line"> <span class="keyword">if</span> h < left:</span><br><span class="line"> cv += left - h</span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> v += cv</span><br><span class="line"> cv = <span class="number">0</span></span><br><span class="line"> left = h</span><br><span class="line"> <span class="keyword">return</span> v</span><br><span class="line"> <span class="keyword">return</span> get_v(h1) + get_v(h2)</span><br></pre></td></tr></table></figure>
<h2 id="71-简化路径"><a href="#71-简化路径" class="headerlink" title="71. 简化路径"></a>71. <a href="https://leetcode-cn.com/problems/simplify-path/" rel="external nofollow" target="_blank">简化路径</a></h2><p>以 Unix 风格给出一个文件的<strong>绝对路径</strong>。右侧不带斜杠<code>/</code>，去除相对路径符号<code>.</code>（当前目录），<code>..</code>（上级目录）。</p>
<figure class="highlight python"><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></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">simplifyPath</span><span class="params">(self, path: str)</span> -> str:</span></span><br><span class="line"> <span class="string">'''</span></span><br><span class="line"><span class="string"> 思路：</span></span><br><span class="line"><span class="string"> 1. 按'/'切分</span></span><br><span class="line"><span class="string"> 2. 去除空字符串</span></span><br><span class="line"><span class="string"> 3. 去除当前目录符号'.'</span></span><br><span class="line"><span class="string"> 4. 抵消'..'前的目录</span></span><br><span class="line"><span class="string"> 5. 给最后输入加入左斜杠'/'</span></span><br><span class="line"><span class="string"> '''</span></span><br><span class="line"> dirs = path.split(<span class="string">'/'</span>)</span><br><span class="line"> <span class="keyword">while</span> <span class="string">''</span> <span class="keyword">in</span> dirs:</span><br><span class="line"> dirs.pop(dirs.index(<span class="string">''</span>))</span><br><span class="line"> <span class="keyword">while</span> <span class="string">'.'</span> <span class="keyword">in</span> dirs:</span><br><span class="line"> dirs.pop(dirs.index(<span class="string">'.'</span>))</span><br><span class="line"> <span class="keyword">while</span> <span class="string">'..'</span> <span class="keyword">in</span> dirs:</span><br><span class="line"> idx = dirs.index(<span class="string">'..'</span>)</span><br><span class="line"> dirs.pop(idx)</span><br><span class="line"> <span class="keyword">if</span> idx > <span class="number">0</span>:</span><br><span class="line"> dirs.pop(idx<span class="number">-1</span>)</span><br><span class="line"> <span class="keyword">return</span> <span class="string">'/'</span> + <span class="string">'/'</span>.join(dirs)</span><br></pre></td></tr></table></figure>
<h1 id="队列"><a href="#队列" class="headerlink" title="队列"></a>队列</h1><p>…</p>
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<div style="text-align:center;color: #ccc;font-size:14px;">-------------本文结束<i class="fa fa-heart"></i>感谢您的阅读-------------</div>
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<p><span>本文标题:</span><a href="/leetcode.html">LeetCode刷题记录(python)</a></p>
<p><span>文章作者:</span><a href="/" title="访问 Keavnn 的个人博客">Keavnn</a></p>
<p><span>发布时间:</span>2020年04月24日 - 11:04</p>
<p><span>最后更新:</span>2020年04月24日 - 14:04</p>
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