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<p>同理，<strong>次对角线上的所有格子的 <code>row + col</code> 是恒定值</strong>。我们可以使用同样的方法，借助数组 <code>diag2</code> 来处理次对角线约束。</p>
<p>根据以上分析，我们便可以写出 <span class="arithmatex">\(n\)</span> 皇后的解题代码。</p>
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<div class="highlight"><span class="filename">n_queens.java</span><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="cm">/* 回溯算法：N 皇后 */</span>
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<a id="__codelineno-9-3" name="__codelineno-9-3" href="#__codelineno-9-3"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">nQueens</span><span class="p">}</span>
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<div class="highlight"><span class="filename">n_queens.dart</span><pre><span></span><code><a id="__codelineno-10-1" name="__codelineno-10-1" href="#__codelineno-10-1"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">backtrack</span><span class="p">}</span>
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<a id="__codelineno-10-3" name="__codelineno-10-3" href="#__codelineno-10-3"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">nQueens</span><span class="p">}</span>
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<h2 id="1231">12.3.1. &nbsp; 复杂度分析<a class="headerlink" href="#1231" title="Permanent link">&para;</a></h2>