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<h1 id="linked-list">Linked List</h1>
<h2 id="21-merge-two-sorted-lists">21. Merge Two Sorted Lists</h2>
<div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">typedef</span> ListNode<span style="color:#f92672">*</span> ListLink;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">Solution</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span><span style="color:#f92672">:</span>
</span></span><span style="display:flex;"><span> ListNode<span style="color:#f92672">*</span> mergeTwoLists(ListNode<span style="color:#f92672">*</span> list1, ListNode<span style="color:#f92672">*</span> list2) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (list1 <span style="color:#f92672">==</span> <span style="color:#66d9ef">nullptr</span>) <span style="color:#66d9ef">return</span> list2;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (list2 <span style="color:#f92672">==</span> <span style="color:#66d9ef">nullptr</span>) <span style="color:#66d9ef">return</span> list1;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#75715e">// 将头节点较小的链表放在 list1
</span></span></span><span style="display:flex;"><span><span style="color:#75715e"></span> <span style="color:#66d9ef">if</span> (list1<span style="color:#f92672">-></span>val <span style="color:#f92672">></span> list2<span style="color:#f92672">-></span>val) swap(list1, list2);
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#75715e">// 遍历 list2，按大小插入 list1
</span></span></span><span style="display:flex;"><span><span style="color:#75715e"></span> <span style="color:#66d9ef">auto</span> ls1 <span style="color:#f92672">=</span> list1, ls2 <span style="color:#f92672">=</span> list2;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (ls2 <span style="color:#f92672">!=</span> <span style="color:#66d9ef">nullptr</span>) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (ls1<span style="color:#f92672">-></span>next <span style="color:#f92672">!=</span> <span style="color:#66d9ef">nullptr</span> and ls2<span style="color:#f92672">-></span>val <span style="color:#f92672">></span> ls1<span style="color:#f92672">-></span>next<span style="color:#f92672">-></span>val) {}
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> { <span style="color:#75715e">// 插入节点
</span></span></span><span style="display:flex;"><span><span style="color:#75715e"></span> ListLink temp <span style="color:#f92672">=</span> <span style="color:#66d9ef">new</span> ListNode(ls2<span style="color:#f92672">-></span>val, ls1<span style="color:#f92672">-></span>next);
</span></span><span style="display:flex;"><span> ls1<span style="color:#f92672">-></span>next <span style="color:#f92672">=</span> temp;
</span></span><span style="display:flex;"><span> ls2 <span style="color:#f92672">=</span> ls2<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> ls1 <span style="color:#f92672">=</span> ls1<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> list1;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div><div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-c" data-lang="c"><span style="display:flex;"><span><span style="color:#66d9ef">typedef</span> ListNode<span style="color:#f92672">*</span> ListLink;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span>class Solution {
</span></span><span style="display:flex;"><span>public:
</span></span><span style="display:flex;"><span> ListNode<span style="color:#f92672">*</span> <span style="color:#a6e22e">mergeTwoLists</span>(ListNode<span style="color:#f92672">*</span> list1, ListNode<span style="color:#f92672">*</span> list2) {
</span></span><span style="display:flex;"><span> ListLink res <span style="color:#f92672">=</span> new <span style="color:#a6e22e">ListNode</span>(<span style="color:#ae81ff">0</span>);
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">auto</span> p <span style="color:#f92672">=</span> res, p1 <span style="color:#f92672">=</span> list1, p2 <span style="color:#f92672">=</span> list2;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (p1 and p2) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (p1<span style="color:#f92672">-></span>val <span style="color:#f92672"><</span> p2<span style="color:#f92672">-></span>val) {
</span></span><span style="display:flex;"><span> p<span style="color:#f92672">-></span>next <span style="color:#f92672">=</span> p1;
</span></span><span style="display:flex;"><span> p1 <span style="color:#f92672">=</span> p1<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> {
</span></span><span style="display:flex;"><span> p<span style="color:#f92672">-></span>next <span style="color:#f92672">=</span> p2;
</span></span><span style="display:flex;"><span> p2 <span style="color:#f92672">=</span> p2<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> p <span style="color:#f92672">=</span> p<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (p1) {
</span></span><span style="display:flex;"><span> p<span style="color:#f92672">-></span>next <span style="color:#f92672">=</span> p1;
</span></span><span style="display:flex;"><span> p1 <span style="color:#f92672">=</span> p1<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> p <span style="color:#f92672">=</span> p<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (p2) {
</span></span><span style="display:flex;"><span> p<span style="color:#f92672">-></span>next <span style="color:#f92672">=</span> p2;
</span></span><span style="display:flex;"><span> p2 <span style="color:#f92672">=</span> p2<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> p <span style="color:#f92672">=</span> p<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> res<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div><div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-java" data-lang="java"><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">Solution</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span>:
</span></span><span style="display:flex;"><span> ListNode<span style="color:#f92672">*</span> mergeTwoLists(ListNode<span style="color:#f92672">*</span> list1, ListNode<span style="color:#f92672">*</span> list2) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (list1 <span style="color:#f92672">==</span> nullptr) <span style="color:#66d9ef">return</span> list2;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (list2 <span style="color:#f92672">==</span> nullptr) <span style="color:#66d9ef">return</span> list1;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (list1<span style="color:#f92672">-></span>val <span style="color:#f92672"><=</span> list2<span style="color:#f92672">-></span>val) {
</span></span><span style="display:flex;"><span> list1<span style="color:#f92672">-></span>next <span style="color:#f92672">=</span> mergeTwoLists(list1<span style="color:#f92672">-></span>next, list2);
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> list1;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> {
</span></span><span style="display:flex;"><span> list2<span style="color:#f92672">-></span>next <span style="color:#f92672">=</span> mergeTwoLists(list1, list2<span style="color:#f92672">-></span>next);
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> list2;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div>
<h3 id="heading">题面</h3>
<p>有两个链表，其中的节点都按照节点值有序排序，请你将这两个链表合并为一个有序链表。</p>
<h3 id="heading-1">题解</h3>
<p><strong>Method One</strong>: 直接法</p>
<ul>
<li>将 <code>head->val</code> 较小的链表作为 <code>list1</code>，放在前面被插入</li>
<li>遍历 <code>list2</code> 按照 <code>node->val</code> 的大小依次插入 <code>list1</code></li>
<li>需特判链表为空的情况</li>
</ul>
<p><strong>Method Two</strong>: 双指针</p>
<ul>
<li>构造一个虚拟头节点 <code>res</code>，并创建指针 <code>p</code></li>
<li>分别为两个链表创建指针 <code>p1</code> 和 <code>p2</code></li>
<li>遍历 <code>p1</code> 和 <code>p2</code>，直到有一个走到终点</li>
<li>将没有走完的指针走完</li>
</ul>
<p><strong>Method Three</strong>: 递归</p>
<ul>
<li>思想和直接法一样，只是插入的逻辑有变化</li>
<li>直接法是始终往一个链表插入</li>
<li>递归法是根据当前节点的 <code>val</code> 来判断插入方向</li>
</ul>
<h2 id="83-remove-duplicates-from-sorted-list">83. Remove Duplicates from Sorted List</h2>
<div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">Solution</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span><span style="color:#f92672">:</span>
</span></span><span style="display:flex;"><span> ListNode<span style="color:#f92672">*</span> deleteDuplicates(ListNode<span style="color:#f92672">*</span> head) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">auto</span> p <span style="color:#f92672">=</span> head;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (p <span style="color:#f92672">!=</span> <span style="color:#66d9ef">nullptr</span> and p<span style="color:#f92672">-></span>next <span style="color:#f92672">!=</span> <span style="color:#66d9ef">nullptr</span>) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (p<span style="color:#f92672">-></span>val <span style="color:#f92672">==</span> p<span style="color:#f92672">-></span>next<span style="color:#f92672">-></span>val) {
</span></span><span style="display:flex;"><span> p<span style="color:#f92672">-></span>next <span style="color:#f92672">=</span> p<span style="color:#f92672">-></span>next<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> p <span style="color:#f92672">=</span> p<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> head;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div>
<h3 id="heading-2">题面</h3>
<p>给你一个有序的链表，请删除其中的重复节点。</p>
<h3 id="heading-3">题解</h3>
<p><strong>Method One</strong>: 直接法</p>
<ul>
<li>因为链表是有序的，所以直接遍历链表</li>
<li>如果 <code>p->val == p->next->val</code> 则删除 <code>p->next</code> 节点</li>
<li>否则继续前进</li>
</ul>
<h2 id="141-linked-list-cycle">141. Linked List Cycle</h2>
<div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">Solution</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span><span style="color:#f92672">:</span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">bool</span> hasCycle(ListNode <span style="color:#f92672">*</span>head) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">auto</span> fast <span style="color:#f92672">=</span> head, slow <span style="color:#f92672">=</span> head;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (fast and fast<span style="color:#f92672">-></span>next) {
</span></span><span style="display:flex;"><span> fast <span style="color:#f92672">=</span> fast<span style="color:#f92672">-></span>next<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> slow <span style="color:#f92672">=</span> slow<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (fast <span style="color:#f92672">==</span> slow) <span style="color:#66d9ef">return</span> true;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> false;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div><div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-c" data-lang="c"><span style="display:flex;"><span>class Solution {
</span></span><span style="display:flex;"><span>public:
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">bool</span> <span style="color:#a6e22e">hasCycle</span>(ListNode <span style="color:#f92672">*</span>head) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (<span style="color:#f92672">!</span>head) <span style="color:#66d9ef">return</span> false;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> cnt <span style="color:#f92672">=</span> <span style="color:#ae81ff">0</span>;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (cnt <span style="color:#f92672"><</span> <span style="color:#ae81ff">10010</span>) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (head<span style="color:#f92672">-></span>next <span style="color:#f92672">==</span> nullptr) <span style="color:#66d9ef">return</span> false;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> head <span style="color:#f92672">=</span> head<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> cnt <span style="color:#f92672">++</span>;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> true;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div>
<h3 id="heading-4">题面</h3>
<p>给你一个链表，请判断链表中是否存在环。</p>
<h3 id="heading-5">题解</h3>
<p><strong>Method One</strong>: 快慢指针</p>
<ul>
<li>若链表中存在环，则遍历链表是一个死循环</li>
<li>因此我们构造两个快慢指针 <code>fast</code> 和 <code>slow</code></li>
<li><code>fast</code> 每次前进 2 步，<code>slow</code> 每次前进 1 步</li>
<li>如果存在环，则 <code>fast</code> 和 <code>slow</code> 会相遇</li>
</ul>
<p><strong>Method Two</strong>: 暴力</p>
<ul>
<li>注意到题目的数据量为 <code>1e4</code></li>
<li>因此我们可以直接前进 <code>10010</code> 步，判断是否走到终点</li>
</ul>
<h2 id="160-intersection-of-two-linked-lists">160. Intersection of Two Linked Lists</h2>
<div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">Solution</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span><span style="color:#f92672">:</span>
</span></span><span style="display:flex;"><span> ListNode <span style="color:#f92672">*</span>getIntersectionNode(ListNode <span style="color:#f92672">*</span>headA, ListNode <span style="color:#f92672">*</span>headB) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">auto</span> p <span style="color:#f92672">=</span> headA, q <span style="color:#f92672">=</span> headB;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (p <span style="color:#f92672">!=</span> q) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (<span style="color:#f92672">!</span>p) p <span style="color:#f92672">=</span> headB;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> p <span style="color:#f92672">=</span> p<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (<span style="color:#f92672">!</span>q) q <span style="color:#f92672">=</span> headA;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> q <span style="color:#f92672">=</span> q<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> p;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div>
<h3 id="heading-6">题面</h3>
<p>有两个链表 <code>headA</code> 和 <code>headB</code>，请你找出它们尾部相同的部分。</p>
<h3 id="heading-7">题解</h3>
<p><strong>Method One</strong>: 双指针</p>
<ul>
<li><code>headA</code> 和 <code>headB</code> 长度不一致，但是尾部相同</li>
<li>则有 <code>headA = preA + tail</code>， <code>headB = preB + tail</code></li>
<li>即 <code>headA + headB = x + tail</code>，<code>headB + headA = y + tail</code></li>
<li>因此直接找 <code>headA + headB</code> 和 <code>headB + headA</code> 的相同尾部</li>
</ul>
<h2 id="203-remove-linked-list-elements">203. Remove Linked List Elements</h2>
<div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">Solution</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span><span style="color:#f92672">:</span>
</span></span><span style="display:flex;"><span> ListNode<span style="color:#f92672">*</span> removeElements(ListNode<span style="color:#f92672">*</span> head, <span style="color:#66d9ef">int</span> val) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (head and head<span style="color:#f92672">-></span>val <span style="color:#f92672">==</span> val) head <span style="color:#f92672">=</span> head<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">auto</span> p <span style="color:#f92672">=</span> head;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (p and p<span style="color:#f92672">-></span>next) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (p<span style="color:#f92672">-></span>next<span style="color:#f92672">-></span>val <span style="color:#f92672">==</span> val) p<span style="color:#f92672">-></span>next <span style="color:#f92672">=</span> p<span style="color:#f92672">-></span>next<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> p <span style="color:#f92672">=</span> p<span style="color:#f92672">-></span>next;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> head;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div>
<h3 id="heading-8">题面</h3>
<p>给你一个链表 <code>head</code> 和一个整数 <code>val</code>，请删除链表中 <code>p->val == val</code> 的节点。</p>
<h3 id="heading-9">题解</h3>
<p><strong>Method One</strong>: 直接法</p>
<ul>
<li>首先特判头节点的值 <code>head->val</code> 是否等于 <code>val</code></li>
<li>然后遍历链表，将 <code>p->val == val</code> 的节点删掉</li>
</ul>
<h1 id="stack">Stack</h1>
<h2 id="20-valid-parentheses">20. Valid Parentheses</h2>
<div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">Solution</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span><span style="color:#f92672">:</span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">bool</span> isValid(string s) {
</span></span><span style="display:flex;"><span> stack<span style="color:#f92672"><</span><span style="color:#66d9ef">char</span><span style="color:#f92672">></span> stk;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">for</span> (<span style="color:#66d9ef">auto</span> c : s) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (stk.size() <span style="color:#f92672">==</span> <span style="color:#ae81ff">0</span>) stk.push(c);
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> <span style="color:#a6e22e">if</span> (stk.top() <span style="color:#f92672">==</span> <span style="color:#e6db74">'('</span> and c <span style="color:#f92672">==</span> <span style="color:#e6db74">')'</span>) stk.pop();
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> <span style="color:#a6e22e">if</span> (stk.top() <span style="color:#f92672">==</span> <span style="color:#e6db74">'['</span> and c <span style="color:#f92672">==</span> <span style="color:#e6db74">']'</span>) stk.pop();
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> <span style="color:#a6e22e">if</span> (stk.top() <span style="color:#f92672">==</span> <span style="color:#960050;background-color:#1e0010">'</span>{<span style="color:#960050;background-color:#1e0010">'</span> and c <span style="color:#f92672">==</span> <span style="color:#e6db74">'}'</span>) stk.pop();
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> stk.push(c);
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> stk.size() <span style="color:#f92672">==</span> <span style="color:#ae81ff">0</span> <span style="color:#f92672">?</span> true <span style="color:#f92672">:</span> false;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div>
<h3 id="heading-10">题面</h3>
<p>给定一个只包括 <code>'('</code>，<code>')'</code>，<code>'{'</code>，<code>'}'</code>，<code>'['</code>，<code>']'</code> 的字符串 <code>s</code> ，判断字符串是否有效。</p>
<p>有效字符串需满足：</p>
<ol>
<li>左括号必须用相同类型的右括号闭合。</li>
<li>左括号必须以正确的顺序闭合。</li>
<li>每个右括号都有一个对应的相同类型的左括号。</li>
</ol>
<h3 id="heading-1-1">题解</h3>
<p><strong>Method One</strong>: 直接法</p>
<ul>
<li>遍历字符串，如果栈为空则入栈</li>
<li>如果栈不为空，则判断栈头和当前字符是否匹配
<ul>
<li>如果匹配，则栈弹出</li>
<li>否则入栈</li>
</ul>
</li>
<li>最终判断栈是否为空，如果为空则合法</li>
</ul>
<h2 id="225-implement-stack-using-queues">225. Implement Stack using Queues</h2>
<div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">MyStack</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">private</span><span style="color:#f92672">:</span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> stk[<span style="color:#ae81ff">200</span>];
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> head;
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span><span style="color:#f92672">:</span>
</span></span><span style="display:flex;"><span> MyStack() <span style="color:#f92672">:</span> head(<span style="color:#f92672">-</span><span style="color:#ae81ff">1</span>) {}
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">void</span> <span style="color:#a6e22e">push</span>(<span style="color:#66d9ef">int</span> x) {
</span></span><span style="display:flex;"><span> stk[<span style="color:#f92672">++</span> head] <span style="color:#f92672">=</span> x;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> <span style="color:#a6e22e">pop</span>() {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> stk[head <span style="color:#f92672">--</span>];
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> <span style="color:#a6e22e">top</span>() {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> stk[head];
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">bool</span> <span style="color:#a6e22e">empty</span>() {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> head <span style="color:#f92672">==</span> <span style="color:#f92672">-</span><span style="color:#ae81ff">1</span>;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div>
<h3 id="heading-2-1">题面</h3>
<p>给定一个只包括 <code>'('</code>，<code>')'</code>，<code>'{'</code>，<code>'}'</code>，<code>'['</code>，<code>']'</code> 的字符串 <code>s</code> ，判断字符串是否有效。</p>
<p>有效字符串需满足：</p>
<ol>
<li>左括号必须用相同类型的右括号闭合。</li>
<li>左括号必须以正确的顺序闭合。</li>
<li>每个右括号都有一个对应的相同类型的左括号。</li>
</ol>
<h3 id="heading-3-1">题解</h3>
<p><strong>Method One</strong>: 数组模拟</p>
<ul>
<li>数据范围 <code>100</code>，则直接开一个长度足够的数组</li>
<li>通过下标 <code>head</code> 来模拟入栈和弹出等</li>
</ul>
<h2 id="496-next-greater-element-i">496. Next Greater Element I</h2>
<div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">Solution</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span><span style="color:#f92672">:</span>
</span></span><span style="display:flex;"><span> vector<span style="color:#f92672"><</span><span style="color:#66d9ef">int</span><span style="color:#f92672">></span> nextGreaterElement(vector<span style="color:#f92672"><</span><span style="color:#66d9ef">int</span><span style="color:#f92672">>&</span> nums1, vector<span style="color:#f92672"><</span><span style="color:#66d9ef">int</span><span style="color:#f92672">>&</span> nums2) {
</span></span><span style="display:flex;"><span> stack<span style="color:#f92672"><</span><span style="color:#66d9ef">int</span><span style="color:#f92672">></span> stk;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> flag[<span style="color:#ae81ff">10010</span>];
</span></span><span style="display:flex;"><span> memset(flag, <span style="color:#f92672">-</span><span style="color:#ae81ff">1</span>, <span style="color:#66d9ef">sizeof</span>(flag));
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">for</span> (<span style="color:#66d9ef">auto</span> i : nums2) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (<span style="color:#f92672">!</span>stk.empty() and stk.top() <span style="color:#f92672"><</span> i) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> x <span style="color:#f92672">=</span> stk.top();
</span></span><span style="display:flex;"><span> flag[x] <span style="color:#f92672">=</span> i;
</span></span><span style="display:flex;"><span> stk.pop();
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> stk.push(i);
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> vector<span style="color:#f92672"><</span><span style="color:#66d9ef">int</span><span style="color:#f92672">></span> res(nums1.size(), <span style="color:#f92672">-</span><span style="color:#ae81ff">1</span>);
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">for</span> (<span style="color:#66d9ef">int</span> i <span style="color:#f92672">=</span> <span style="color:#ae81ff">0</span>, l <span style="color:#f92672">=</span> nums1.size(); i <span style="color:#f92672"><</span> l; i <span style="color:#f92672">++</span>) {
</span></span><span style="display:flex;"><span> res[i] <span style="color:#f92672">=</span> flag[nums1[i]];
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> res;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div>
<h3 id="heading-4-1">题面</h3>
<p><code>nums1</code> 中数字 <code>x</code> 的 <strong>下一个更大元素</strong> 是指 <code>x</code> 在 <code>nums2</code> 中对应位置 <strong>右侧</strong> 的 <strong>第一个</strong> 比 <code>x</code> 大的元素。</p>
<p>给你两个 <strong>没有重复元素</strong> 的数组 <code>nums1</code> 和 <code>nums2</code> ，下标从 <strong>0</strong> 开始计数，其中<code>nums1</code> 是 <code>nums2</code> 的子集。</p>
<p>对于每个 <code>0 <= i < nums1.length</code> ，找出满足 <code>nums1[i] == nums2[j]</code> 的下标 <code>j</code> ，并且在 <code>nums2</code> 确定 <code>nums2[j]</code> 的 <strong>下一个更大元素</strong> 。如果不存在下一个更大元素，那么本次查询的答案是 <code>-1</code> 。</p>
<p>返回一个长度为 <code>nums1.length</code> 的数组 <code>ans</code> 作为答案，满足 <code>ans[i]</code> 是如上所述的 <strong>下一个更大元素</strong> 。</p>
<h3 id="heading-5-1">题解</h3>
<p><strong>Method One</strong>: 单调栈</p>
<ul>
<li>我们需要在 <code>nums2</code> 中找到 <code>x</code> 右侧比 <code>x</code> 大的数字 <code>y</code></li>
<li>构造一个单调栈，遍历 <code>nums2</code>，记录被弹出的元素的值 <code>x</code> 和新插入的 <code>y</code></li>
<li>这些元素会被弹出，说明 <code>x</code> 的右侧存在比 <code>x</code> 大的 <code>y</code></li>
<li>最后遍历 <code>nums1</code> 找到每个 <code>x</code> 对应的 <code>y</code></li>
</ul>
<p>注意：需注意数据范围 <code>[0, 1e4]</code> 开足够长的数组和下标的记录。</p>
<h2 id="682-baseball-game">682. Baseball Game</h2>
<div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">Solution</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span><span style="color:#f92672">:</span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> calPoints(vector<span style="color:#f92672"><</span>string<span style="color:#f92672">>&</span> operations) {
</span></span><span style="display:flex;"><span> stack<span style="color:#f92672"><</span><span style="color:#66d9ef">int</span><span style="color:#f92672">></span> stk;
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">for</span> (<span style="color:#66d9ef">auto</span> s : operations) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (s <span style="color:#f92672">==</span> <span style="color:#e6db74">"+"</span>) {
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> x <span style="color:#f92672">=</span> stk.top(); stk.pop();
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> y <span style="color:#f92672">=</span> stk.top(); stk.pop();
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> stk.push(y); stk.push(x); stk.push(x <span style="color:#f92672">+</span> y);
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> <span style="color:#a6e22e">if</span> (s <span style="color:#f92672">==</span> <span style="color:#e6db74">"D"</span>) stk.push(stk.top() <span style="color:#f92672">*</span> <span style="color:#ae81ff">2</span>);
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> <span style="color:#a6e22e">if</span> (s <span style="color:#f92672">==</span> <span style="color:#e6db74">"C"</span>) stk.pop();
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">else</span> stk.push(std<span style="color:#f92672">::</span>stoi(s));
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> sum <span style="color:#f92672">=</span> <span style="color:#ae81ff">0</span>;
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">while</span> (<span style="color:#f92672">!</span>stk.empty()) {
</span></span><span style="display:flex;"><span> sum <span style="color:#f92672">+=</span> stk.top();
</span></span><span style="display:flex;"><span> stk.pop();
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>
</span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> sum;
</span></span><span style="display:flex;"><span> }
</span></span><span style="display:flex;"><span>};
</span></span></code></pre></div>
<h3 id="heading-6-1">题面</h3>
<p>你现在是一场采用特殊赛制棒球比赛的记录员。这场比赛由若干回合组成，过去几回合的得分可能会影响以后几回合的得分。</p>
<p>比赛开始时，记录是空白的。你会得到一个记录操作的字符串列表 <code>ops</code>，其中 <code>ops[i]</code> 是你需要记录的第 <code>i</code> 项操作，<code>ops</code> 遵循下述规则：</p>
<ol>
<li>整数 <code>x</code> - 表示本回合新获得分数 <code>x</code></li>
<li><code>"+"</code> - 表示本回合新获得的得分是前两次得分的总和。题目数据保证记录此操作时前面总是存在两个有效的分数。</li>
<li><code>"D"</code> - 表示本回合新获得的得分是前一次得分的两倍。题目数据保证记录此操作时前面总是存在一个有效的分数。</li>
<li><code>"C"</code> - 表示前一次得分无效，将其从记录中移除。题目数据保证记录此操作时前面总是存在一个有效的分数。</li>
</ol>
<p>请你返回记录中所有得分的总和。</p>
<h3 id="heading-7-1">题解</h3>
<p><strong>Method One</strong>: 直接法</p>
<ul>
<li>用栈模拟题面的操作</li>
</ul>
<h2 id="844-backspace-string-compare">844. Backspace String Compare</h2>
<div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">class</span> <span style="color:#a6e22e">Solution</span> {
</span></span><span style="display:flex;"><span><span style="color:#66d9ef">public</span><span style="color:#f92672">:</span>