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chapter_dynamic_programming/intro_to_dynamic_programming/index.html
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<span class="md-ellipsis">
第 2 章 &nbsp; 复杂度
第 2 章 &nbsp; 时空复杂度
</span>
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<nav class="md-nav" data-md-level="1" aria-labelledby="__nav_3_label" aria-expanded="false">
<label class="md-nav__title" for="__nav_3">
<span class="md-nav__icon md-icon"></span>
第 2 章 &nbsp; 复杂度
第 2 章 &nbsp; 时空复杂度
</label>
<ul class="md-nav__list" data-md-scrollfix>
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<span class="md-ellipsis">
第 6 章 &nbsp; 散列表
第 6 章 &nbsp; 哈希表
</span>
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<nav class="md-nav" data-md-level="1" aria-labelledby="__nav_7_label" aria-expanded="false">
<label class="md-nav__title" for="__nav_7">
<span class="md-nav__icon md-icon"></span>
第 6 章 &nbsp; 散列表
第 6 章 &nbsp; 哈希表
</label>
<ul class="md-nav__list" data-md-scrollfix>
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<p>回溯算法通常并不显式地对问题进行拆解，而是将问题看作一系列决策步骤，通过试探和剪枝，搜索所有可能的解。</p>
<p>我们可以尝试从问题分解的角度分析这道题。设爬到第 <span class="arithmatex">\(i\)</span> 阶共有 <span class="arithmatex">\(dp[i]\)</span> 种方案，那么 <span class="arithmatex">\(dp[i]\)</span> 就是原问题，其子问题包括:</p>
<div class="arithmatex">\[
dp[i-1] , dp[i-2] , \cdots , dp[2] , dp[1]
dp[i-1] , dp[i-2] , \dots , dp[2] , dp[1]
\]</div>
<p>由于每轮只能上 <span class="arithmatex">\(1\)</span> 阶或 <span class="arithmatex">\(2\)</span> 阶，因此当我们站在第 <span class="arithmatex">\(i\)</span> 阶楼梯上时，上一轮只可能站在第 <span class="arithmatex">\(i - 1\)</span> 阶或第 <span class="arithmatex">\(i - 2\)</span> 阶上。换句话说，我们只能从第 <span class="arithmatex">\(i -1\)</span> 阶或第 <span class="arithmatex">\(i - 2\)</span> 阶前往第 <span class="arithmatex">\(i\)</span> 阶。</p>
<p>由此便可得出一个重要推论：<strong>爬到第 <span class="arithmatex">\(i - 1\)</span> 阶的方案数加上爬到第 <span class="arithmatex">\(i - 2\)</span> 阶的方案数就等于爬到第 <span class="arithmatex">\(i\)</span> 阶的方案数</strong>。公式如下：</p>