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<h2 class="post-title" itemprop="name headline">Leetcode 841: Keys and Rooms
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<p><strong>难度：Medium</strong></p>
<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>有N个房间，开始时你位于0号房间。每个房间有不同的号码：0，1，2，…，N-1，并且房间里可能有一些钥匙能使你进入下一个房间。</p>
<p>在形式上，对于每个房间i都有一个钥匙列表rooms[i]，每个钥匙rooms[i][j]由[0,1，…，N-1]中的一个整数表示，其中N=rooms.length。 钥匙rooms[i][j]=v可以打开编号为v的房间。</p>
<a id="more"></a>
<p>最初，除0号房间外的其余所有房间都被锁住。</p>
<p>你可以自由地在房间之间来回走动。</p>
<p>如果能进入每个房间返回true，否则返回false。</p>
<h3 id="示例"><a href="#示例" class="headerlink" title="示例"></a>示例</h3><h4 id="示例1"><a href="#示例1" class="headerlink" title="示例1"></a>示例1</h4><figure class="highlight shell"><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">输入: [[1],[2],[3],[]]</span><br><span class="line">输出: true</span><br><span class="line">解释: </span><br><span class="line">我们从0号房间开始，拿到钥匙1。</span><br><span class="line">之后我们去1号房间，拿到钥匙2。</span><br><span class="line">然后我们去2号房间，拿到钥匙3。</span><br><span class="line">最后我们去了3号房间。</span><br><span class="line">由于我们能够进入每个房间，我们返回true。</span><br></pre></td></tr></table></figure>
<h4 id="示例2"><a href="#示例2" class="headerlink" title="示例2"></a>示例2</h4><figure class="highlight shell"><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：[[1,3],[3,0,1],[2],[0]]</span><br><span class="line">输出：false</span><br><span class="line">解释：我们不能进入2号房间。</span><br></pre></td></tr></table></figure>
<h3 id="提示"><a href="#提示" class="headerlink" title="提示"></a>提示</h3><ul>
<li>1 <= rooms.length <= 1000</li>
<li>0 <= rooms[i].length <= 1000</li>
<li>所有房间中的钥匙数量总计不超过3000。</li>
</ul>
<h3 id="解题思路"><a href="#解题思路" class="headerlink" title="解题思路"></a>解题思路</h3><p>这个题目其实整体来看是一个图，房间就是图中的点，钥匙就是一个点到另一个点的有向边。最后就是找出哪个点没有访问到。这个我们很容易想到用深度优先遍历来做。<br>具体参见代码。</p>
<h3 id="解题代码"><a href="#解题代码" class="headerlink" title="解题代码"></a>解题代码</h3><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></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">public</span> <span class="keyword">boolean</span> <span class="title">canVisitAllRooms</span><span class="params">(List<List<Integer>> rooms)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> n = rooms.size();</span><br><span class="line"> <span class="keyword">boolean</span>[] visited = <span class="keyword">new</span> <span class="keyword">boolean</span>[n];</span><br><span class="line"> dfs(rooms, <span class="number">0</span>, visited);</span><br><span class="line"> visited[<span class="number">0</span>] = <span class="keyword">true</span>;</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">boolean</span> ele : visited) {</span><br><span class="line"> <span class="keyword">if</span> (!ele) {</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">false</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">true</span>;</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="function"><span class="keyword">public</span> <span class="keyword">void</span> <span class="title">dfs</span><span class="params">(List<List<Integer>> rooms, <span class="keyword">int</span> r, <span class="keyword">boolean</span>[] visited)</span> </span>{</span><br><span class="line"> <span class="keyword">for</span> (Integer ele : rooms.get(r)) {</span><br><span class="line"> <span class="keyword">if</span> (!visited[ele]) {</span><br><span class="line"> visited[ele] = <span class="keyword">true</span>;</span><br><span class="line"> dfs(rooms, ele, visited);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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