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<title>12.4.   N 皇后问题 - Hello 算法</title>
<title>13.4.   N 皇后问题 - Hello 算法</title>
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12.4. &nbsp; N 皇后问题
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<h1 id="124-n">12.4. &nbsp; N 皇后问题<a class="headerlink" href="#124-n" title="Permanent link">&para;</a></h1>
<h1 id="134-n">13.4. &nbsp; N 皇后问题<a class="headerlink" href="#134-n" title="Permanent link">&para;</a></h1>
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<p>根据国际象棋的规则，皇后可以攻击与之处在同一行或同一列或同一斜线上的棋子。给定 <span class="arithmatex">\(n\)</span> 个皇后和一个 <span class="arithmatex">\(n \times n\)</span> 大小的棋盘，寻找使得所有皇后之间无法相互攻击的摆放方案。</p>
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<h2 id="1341">13.4.1. &nbsp; 复杂度分析<a class="headerlink" href="#1341" title="Permanent link">&para;</a></h2>
<p>逐行放置 <span class="arithmatex">\(n\)</span> 次，考虑列约束，则从第一行到最后一行分别有 <span class="arithmatex">\(n, n-1, \cdots, 2, 1\)</span> 个选择，<strong>因此时间复杂度为 <span class="arithmatex">\(O(n!)\)</span></strong> 。实际上，根据对角线约束的剪枝也能够大幅地缩小搜索空间，因而搜索效率往往优于以上时间复杂度。</p>
<p><code>state</code> 使用 <span class="arithmatex">\(O(n^2)\)</span> 空间，<code>cols</code> , <code>diags1</code> , <code>diags2</code> 皆使用 <span class="arithmatex">\(O(n)\)</span> 空间。最大递归深度为 <span class="arithmatex">\(n\)</span> ，使用 <span class="arithmatex">\(O(n)\)</span> 栈帧空间。因此，<strong>空间复杂度为 <span class="arithmatex">\(O(n^2)\)</span></strong> 。</p>
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