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<a href="/2021/02/19/algorithm/Minimum-Number-of-Refueling-Stops/" class="post-title-link" itemprop="url">从Minimum Number of Refueling Stops谈DP降维</a>
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<time title="Created: 2021-02-19 22:49:40 / Modified: 23:35:02" itemprop="dateCreated datePublished" datetime="2021-02-19T22:49:40+09:00">2021-02-19</time>
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<h4 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h4><p><a href="https://leetcode.com/problems/minimum-number-of-refueling-stops/" target="_blank" rel="noopener">Link</a></p>
<p>给定一个<code>target</code>和<code>stations</code>,求最少加几次油可以到达目的地。</p>
<h4 id="Answer"><a href="#Answer" class="headerlink" title="Answer"></a>Answer</h4><p>这是一道DP题</p>
<p><code>dp[i][j]</code>: 经过<code>i</code>个加油站加<code>j</code>次油最多能走多少公里。这里有个限制<code>j <= i</code></p>
<p>当经过第<code>i</code>个加油站有两个选择，只有加油或者不加油两种选择。</p>
<ol>
<li>如果不加油的话, <code>dp[i][j] = dp[i-1][j]</code></li>
<li>如果加油的话，<code>dp[i][j] = dp[i-1][j-1] + station[i-1][1]</code></li>
</ol>
<p>最后我们遍历<code>j</code>, 代表经过所有车站最多加<code>j</code>次油可以跑多少公里。如果大于等于<code>target</code>那么就返回<code>j</code>. 如果没有结果的话就返回<code>-1</code>。</p>
<figure class="highlight c"><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="class"><span class="keyword">class</span> <span class="title">Solution</span> {</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line"> <span class="function"><span class="keyword">int</span> <span class="title">minRefuelStops</span><span class="params">(<span class="keyword">int</span> target, <span class="keyword">int</span> startFuel, <span class="built_in">vector</span><<span class="built_in">vector</span><<span class="keyword">int</span>>>& s)</span> </span>{</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">int</span> n = s.size();</span><br><span class="line"> <span class="comment">// dp[i][j] 经过第i个加油站，加j次油，可以跑的最远距离</span></span><br><span class="line"> <span class="built_in">vector</span><<span class="built_in">vector</span><<span class="keyword">long</span>>> dp(n+<span class="number">1</span>, <span class="built_in">vector</span><<span class="keyword">long</span>>(n+<span class="number">1</span>, <span class="number">0</span>));</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>; i <= n; ++i) {</span><br><span class="line"> dp[i][<span class="number">0</span>] = startFuel;</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">1</span>; i <= n; ++i) {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j = <span class="number">1</span>; j<= i; ++j) {</span><br><span class="line"> <span class="comment">// 不加油</span></span><br><span class="line"> <span class="keyword">if</span> (dp[i<span class="number">-1</span>][j] >= s[i<span class="number">-1</span>][<span class="number">0</span>]) {</span><br><span class="line"> dp[i][j] = dp[i<span class="number">-1</span>][j];</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 加油</span></span><br><span class="line"> <span class="keyword">if</span> (dp[i<span class="number">-1</span>][j<span class="number">-1</span>] >= s[i<span class="number">-1</span>][<span class="number">0</span>]) {</span><br><span class="line"> dp[i][j] = max(dp[i][j], dp[i<span class="number">-1</span>][j<span class="number">-1</span>] + s[i<span class="number">-1</span>][<span class="number">1</span>]);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>; i <= n; ++i) {</span><br><span class="line"> <span class="keyword">if</span> (dp[n][i] >= target) <span class="keyword">return</span> i;</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">return</span> <span class="number">-1</span>;</span><br><span class="line"> }</span><br><span class="line">};</span><br></pre></td></tr></table></figure>
<h4 id="Follow-Up"><a href="#Follow-Up" class="headerlink" title="Follow Up"></a>Follow Up</h4><p>这题用了二维数组，我们观察到<code>dp</code>在更新的时候，只跟前有限个状态有关 </p>
<ul>
<li><p><code>dp[i][j] <=> dp[i-1][j]</code></p>
</li>
<li><p><code>dp[i][j] <=> dp[i-1][j-1]</code></p>
</li>
</ul>
<p>我们看到<code>dp[i][*]</code>总是和前一个状态<code>dp[i-1][*]</code>有关，因此我们可以降低一个维度，也就是把第一维也删去。变为</p>
<ul>
<li><p><code>dp[j] <=> dp[j]</code></p>
</li>
<li><p><code>dp[j] <=> dp[j-1]</code></p>
</li>
</ul>
<p>但是我们要注意<code>dp[j] = dp[j-1]</code>. 如果按顺序更新的话<code>dp[1] <=> dp[0]</code>… <code>dp[2] <=> dp[1]</code>. 在更新<code>dp[2]</code>的时候使用的<code>dp[1]</code>是本轮更新过的，会导致答案错误，因此我们应该尝试倒序更新。<code>dp[2] <=> dp[1]</code>…<code>dp[1] <=> dp[0]</code>. </p>
<figure class="highlight c"><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></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="keyword">public</span>:</span><br><span class="line"> <span class="function"><span class="keyword">int</span> <span class="title">minRefuelStops</span><span class="params">(<span class="keyword">int</span> target, <span class="keyword">int</span> startFuel, <span class="built_in">vector</span><<span class="built_in">vector</span><<span class="keyword">int</span>>>& s)</span> </span>{</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">int</span> n = s.size();</span><br><span class="line"> <span class="comment">// dp[i] 加i次油，可以跑的最远距离</span></span><br><span class="line"> <span class="built_in">vector</span><<span class="keyword">long</span>> dp(n+<span class="number">1</span>);</span><br><span class="line"> dp[<span class="number">0</span>] = startFuel;</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">1</span>; i <= n; ++i) {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j = i; j >= <span class="number">1</span>; --j) {</span><br><span class="line"> <span class="comment">// 不加油</span></span><br><span class="line"> <span class="keyword">if</span> (dp[j] >= s[i<span class="number">-1</span>][<span class="number">0</span>]) {</span><br><span class="line"> dp[j] = dp[j];</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 加油</span></span><br><span class="line"> <span class="keyword">if</span> (dp[j<span class="number">-1</span>] >= s[i<span class="number">-1</span>][<span class="number">0</span>]) {</span><br><span class="line"> dp[j] = max(dp[j], dp[j<span class="number">-1</span>] + s[i<span class="number">-1</span>][<span class="number">1</span>]);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>; i <= n; ++i) {</span><br><span class="line"> <span class="keyword">if</span> (dp[i] >= target) <span class="keyword">return</span> i;</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">return</span> <span class="number">-1</span>;</span><br><span class="line"> }</span><br><span class="line">};</span><br></pre></td></tr></table></figure>
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<a href="/2021/02/14/leetcode/Domino-and-Tromino-Tiling/" class="post-title-link" itemprop="url">Domino and Tromino Tiling</a>
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<h4 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h4><p><a href="https://leetcode.com/problems/domino-and-tromino-tiling/" target="_blank" rel="noopener">Link</a></p>
<p>很有趣的一道题，给定一个<code>2*N</code>的格子，一共有2种形状，问有几种拼法。</p>
<h4 id="Answer"><a href="#Answer" class="headerlink" title="Answer"></a>Answer</h4><p>虽然只有2种形状，一种是<code>domino</code>, 一种是<code>tromino</code>, 但是由于可以旋转，所以拼的时候一共有<code>domino</code>2种+<code>tromino</code>4种=6种。最开始也是想用<code>dp</code>做。但是对于<code>tromino</code>就不适用了。论坛有个人定一个三个<code>dp</code>来做。<br><a href="https://leetcode.com/problems/domino-and-tromino-tiling/" target="_blank" rel="noopener">https://leetcode.com/problems/domino-and-tromino-tiling/</a></p>
<p>时间: $O(N)$<br>空间: $O(3N)$<br>由于只需要最多前两种的状态，所以我们也可以用常数来保存状态使得空间复杂度只有$O(1)$</p>
<figure class="highlight c"><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></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="keyword">public</span>:</span><br><span class="line"> <span class="function"><span class="keyword">int</span> <span class="title">numTilings</span><span class="params">(<span class="keyword">int</span> N)</span> </span>{</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">if</span> (N == <span class="number">1</span>) <span class="keyword">return</span> <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">int</span> mod = <span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line"> <span class="built_in">vector</span><<span class="keyword">long</span>> dp(N+<span class="number">1</span>, <span class="number">0</span>);</span><br><span class="line"> <span class="built_in">vector</span><<span class="keyword">long</span>> dpUp(N+<span class="number">1</span>, <span class="number">0</span>);</span><br><span class="line"> <span class="built_in">vector</span><<span class="keyword">long</span>> dpDown(N+<span class="number">1</span>, <span class="number">0</span>);</span><br><span class="line"> </span><br><span class="line"> dp[<span class="number">1</span>] = <span class="number">1</span>;</span><br><span class="line"> </span><br><span class="line"> dp[<span class="number">2</span>] = <span class="number">2</span>;</span><br><span class="line"> dpUp[<span class="number">2</span>] = <span class="number">1</span>;</span><br><span class="line"> dpDown[<span class="number">2</span>] = <span class="number">1</span>;</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">3</span>; i <= N; ++i) {</span><br><span class="line"> dp[i] = (dp[i<span class="number">-1</span>] + dp[i<span class="number">-2</span>] + dpUp[i<span class="number">-1</span>] + dpDown[i<span class="number">-1</span>]) % mod;</span><br><span class="line"> </span><br><span class="line"> dpUp[i] = (dp[i<span class="number">-2</span>] + dpDown[i<span class="number">-1</span>]) % mod;</span><br><span class="line"> </span><br><span class="line"> dpDown[i] = (dp[i<span class="number">-2</span>] + dpUp[i<span class="number">-1</span>]) % mod;</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">return</span> dp[N];</span><br><span class="line"> }</span><br><span class="line">};</span><br></pre></td></tr></table></figure>
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<h4 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h4><p><a href="https://leetcode.com/problems/3sum-with-multiplicity/" target="_blank" rel="noopener">Link</a></p>
<p><code>3Sum</code>的升级版，给定一个数列，然后求所有加起来和是<code>target</code>的子数列的个数。</p>
<h4 id="Answer"><a href="#Answer" class="headerlink" title="Answer"></a>Answer</h4><p>由于答案不是单一的，肯定根每个数字出现的次数有关，所以用一个<code>map</code>来储存所有数字的个数。然后我们分3种情况。假设<code>nums[i]+nums[j]+nums[k]=target</code></p>
<ol>
<li><p>假设三个数字都是一样的。即<code>nums[i] == nums[j] == nums[k]</code>, 那么就是组合问题，从<code>N</code>个数字里面挑出3个。</p>
</li>
<li><p>假设两个数字一样。即<code>nums[i] == nums[j] != nums[k]</code>, 也就是从<code>N</code>个数字里面挑出2个乘以<code>nums[k]</code>出现的次数。</p>
</li>
<li><p>假设三个数字都不一样。这种比较复杂。嫁入给定的数组是<code>[1,2,3]</code>, <code>target=6</code>, 如果我们双循环数组，我们会得到<code>[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]</code>. 但是其实他们只能算一种。所以我们应该做个简单限制，当<code>nums[i] < nums[j] < nums[k]</code>或者<code>nums[i] > nums[j] > nums[k]</code>时候，我们才计算。</p>
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<p>时间复杂度: $O(N^2)$<br>空间复杂度: $O(N)$</p>
<figure class="highlight c"><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="class"><span class="keyword">class</span> <span class="title">Solution</span> {</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line"> <span class="function"><span class="keyword">int</span> <span class="title">threeSumMulti</span><span class="params">(<span class="built_in">vector</span><<span class="keyword">int</span>>& A, <span class="keyword">int</span> target)</span> </span>{</span><br><span class="line"> </span><br><span class="line"> <span class="built_in">unordered_map</span><<span class="keyword">int</span>,<span class="keyword">long</span>> m;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">auto</span> c: A) {</span><br><span class="line"> m[c]++;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">long</span> res = <span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">auto</span> it1: m){</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">auto</span> it2: m) {</span><br><span class="line"> <span class="keyword">int</span> i = it1.first, j = it2.first, k = target - i - j;</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">if</span> (!m.count(k)) <span class="keyword">continue</span>;</span><br><span class="line"> <span class="keyword">if</span> (i == j && j == k) {</span><br><span class="line"> res += m[i] * (m[i]<span class="number">-1</span>) * (m[i]<span class="number">-2</span>) / <span class="number">6</span>;</span><br><span class="line"> }<span class="keyword">else</span> <span class="keyword">if</span> (i == j && i != k) {</span><br><span class="line"> res += m[i] * (m[i]<span class="number">-1</span>) / <span class="number">2</span> * m[k];</span><br><span class="line"> }<span class="keyword">else</span> <span class="keyword">if</span> (i < j && j < k) {</span><br><span class="line"> res += m[i] * m[j] * m[k];</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> res % <span class="keyword">int</span>(<span class="number">1e9</span>+<span class="number">7</span>);</span><br><span class="line"> }</span><br><span class="line">};</span><br></pre></td></tr></table></figure>
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