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ComputerScience/笔记/MarsCode 难题记录/index.html renamed to ComputerScience/笔记/MarsCode 错题本/index.html

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<meta name="description" content="51 和的逆运算 这题在给定的和不重复的情况下很简单： 首先升序排序好数组sums，生成答案数组nums[n]。 nums[0] + nums[1] 必然等于sums[0]（最小值），nums[0] + nums[2] 必然等于sums[1]（次小值）, … , nums[n-2] + nums[n-1] 必然等于sums[lastIndex]（最大值）。 可以反向推测出nums[0] &#x3D; (">
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<meta property="og:description" content="51 和的逆运算 这题在给定的和不重复的情况下很简单： 首先升序排序好数组sums，生成答案数组nums[n]。 nums[0] + nums[1] 必然等于sums[0]（最小值），nums[0] + nums[2] 必然等于sums[1]（次小值）, … , nums[n-2] + nums[n-1] 必然等于sums[lastIndex]（最大值）。 可以反向推测出nums[0] &#x3D; (">
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<content><![CDATA[<h1 id="51-和的逆运算"><a class="markdownIt-Anchor" href="#51-和的逆运算"></a> 51 和的逆运算</h1>
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<p>这题在给定的和<strong>不重复</strong>的情况下很简单：</p>
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<p>找到这个有序两两和序列，就可以根据上面的算法反推出原来的数组。</p>
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<p>我不知道有没有一种数学算法可以优雅地找到有序两两和序列sums，一举这个问题。但是全排列一定能找到这个sums。以下是第51题的全部代码：</p>
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<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">import</span> java.util.Arrays;</span><br><span class="line"></span><br><span class="line"><span class="keyword">public</span> <span class="keyword">class</span> <span class="title class_">Main</span> &#123;</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">static</span> String ans;</span><br><span class="line"></span><br><span class="line"> <span class="keyword">public</span> <span class="keyword">static</span> String <span class="title function_">solution</span><span class="params">(<span class="type">int</span> n, <span class="type">int</span>[] sums)</span> &#123;</span><br><span class="line"> ans = <span class="string">&quot;Impossible&quot;</span>;</span><br><span class="line"> fullArrange(sums, <span class="number">0</span>, n);</span><br><span class="line"> <span class="comment">// System.out.println(ans);</span></span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line"> &#125;</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">private</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title function_">check</span><span class="params">(<span class="type">int</span> n, <span class="type">int</span>[] sums)</span> &#123;</span><br><span class="line"> <span class="type">int</span>[] nums = <span class="keyword">new</span> <span class="title class_">int</span>[n];</span><br><span class="line"> nums[<span class="number">0</span>] = (<span class="type">int</span>) Math.round((sums[<span class="number">0</span>] + sums[<span class="number">1</span>] - sums[n-<span class="number">1</span>]) / <span class="number">2.0</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">1</span>; i &lt; n; i += <span class="number">1</span>) &#123;</span><br><span class="line"> nums[i] = sums[i-<span class="number">1</span>] - nums[<span class="number">0</span>];</span><br><span class="line"> &#125;</span><br><span class="line"> </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>, k = <span class="number">0</span>; i &lt; n; i += <span class="number">1</span>) &#123;</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> <span class="variable">j</span> <span class="operator">=</span> i+<span class="number">1</span>; j &lt; n; j += <span class="number">1</span>) &#123;</span><br><span class="line"> <span class="keyword">if</span> (nums[i] + nums[j] != sums[k++]) &#123;</span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line"> &#125;</span><br><span class="line"> &#125;</span><br><span class="line"> &#125;</span><br><span class="line"> </span><br><span class="line"> <span class="type">StringBuilder</span> <span class="variable">result</span> <span class="operator">=</span> <span class="keyword">new</span> <span class="title class_">StringBuilder</span>();</span><br><span class="line"> Arrays.sort(nums);</span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> num: nums) &#123;</span><br><span class="line"> result.append(num + <span class="string">&quot; &quot;</span>);</span><br><span class="line"> &#125;</span><br><span class="line"> ans = <span class="keyword">new</span> <span class="title class_">String</span>(result.deleteCharAt(result.length()-<span class="number">1</span>));</span><br><span class="line"> &#125;</span><br><span class="line"></span><br><span class="line"> <span class="comment">// 全排列，确定sums第k位的值</span></span><br><span class="line"> <span class="keyword">private</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title function_">fullArrange</span><span class="params">(<span class="type">int</span>[] sums, <span class="type">int</span> k, <span class="type">int</span> n)</span> &#123;</span><br><span class="line"> <span class="keyword">if</span> (k == sums.length) &#123;</span><br><span class="line"> check(n, sums);</span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line"> &#125;</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">for</span> (<span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> k, l = sums.length; i &lt; l; i += <span class="number">1</span>) &#123;</span><br><span class="line"> swap(sums, i, k);</span><br><span class="line"> fullArrange(sums, k + <span class="number">1</span>, n);</span><br><span class="line"> swap(sums, i, k);</span><br><span class="line"> &#125;</span><br><span class="line"> &#125;</span><br><span class="line"> </span><br><span class="line"> <span class="keyword">private</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title function_">swap</span><span class="params">(<span class="type">int</span>[] a, <span class="type">int</span> i, <span class="type">int</span> j)</span> &#123;</span><br><span class="line"> <span class="type">int</span> <span class="variable">t</span> <span class="operator">=</span> a[i];</span><br><span class="line"> a[i] = a[j];</span><br><span class="line"> a[j] = t;</span><br><span class="line"> &#125;</span><br><span class="line"></span><br><span class="line"> <span class="keyword">public</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title function_">main</span><span class="params">(String[] args)</span> &#123;</span><br><span class="line"> <span class="comment">// You can add more test cases here</span></span><br><span class="line"> <span class="type">int</span>[] sums1 = &#123;<span class="number">1269</span>, <span class="number">1160</span>, <span class="number">1663</span>&#125;;</span><br><span class="line"> <span class="type">int</span>[] sums2 = &#123;<span class="number">1</span>, <span class="number">1</span>, <span class="number">1</span>&#125;;</span><br><span class="line"> <span class="type">int</span>[] sums3 = &#123;<span class="number">226</span>, <span class="number">223</span>, <span class="number">225</span>, <span class="number">224</span>, <span class="number">227</span>, <span class="number">229</span>, <span class="number">228</span>, <span class="number">226</span>, <span class="number">225</span>, <span class="number">227</span>&#125;;</span><br><span class="line"> <span class="type">int</span>[] sums4 = &#123;-<span class="number">1</span>, <span class="number">0</span>, -<span class="number">1</span>, -<span class="number">2</span>, <span class="number">1</span>, <span class="number">0</span>, -<span class="number">1</span>, <span class="number">1</span>, <span class="number">0</span>, -<span class="number">1</span>&#125;;</span><br><span class="line"> <span class="type">int</span>[] sums5 = &#123;<span class="number">79950</span>, <span class="number">79936</span>, <span class="number">79942</span>, <span class="number">79962</span>, <span class="number">79954</span>, <span class="number">79972</span>, <span class="number">79960</span>, <span class="number">79968</span>, <span class="number">79924</span>, <span class="number">79932</span>&#125;;</span><br><span class="line"></span><br><span class="line"> System.out.println(solution(<span class="number">3</span>, sums1).equals(<span class="string">&quot;383 777 886&quot;</span>));</span><br><span class="line"> System.out.println(solution(<span class="number">3</span>, sums2).equals(<span class="string">&quot;Impossible&quot;</span>));</span><br><span class="line"> System.out.println(solution(<span class="number">5</span>, sums3).equals(<span class="string">&quot;111 112 113 114 115&quot;</span>));</span><br><span class="line"> System.out.println(solution(<span class="number">5</span>, sums4).equals(<span class="string">&quot;-1 -1 0 0 1&quot;</span>));</span><br><span class="line"> System.out.println(solution(<span class="number">5</span>, sums5).equals(<span class="string">&quot;39953 39971 39979 39983 39989&quot;</span>));</span><br><span class="line"> &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>]]></content>
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<tags>
6659+
<tag>笔记</tag>
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</tags>
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</entry>
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</search>

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