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@ -3115,22 +3115,22 @@
<ul class="md-nav__list" data-md-component="toc" data-md-scrollfix>
<li class="md-nav__item">
<a href="#_1" class="md-nav__link">
贪心策略确定
<a href="#1" class="md-nav__link">
1. &nbsp; 贪心策略确定
</a>
</li>
<li class="md-nav__item">
<a href="#_2" class="md-nav__link">
代码实现
<a href="#2" class="md-nav__link">
2. &nbsp; 代码实现
</a>
</li>
<li class="md-nav__item">
<a href="#_3" class="md-nav__link">
正确性证明
<a href="#3" class="md-nav__link">
3. &nbsp; 正确性证明
</a>
</li>
@ -3390,22 +3390,22 @@
<ul class="md-nav__list" data-md-component="toc" data-md-scrollfix>
<li class="md-nav__item">
<a href="#_1" class="md-nav__link">
贪心策略确定
<a href="#1" class="md-nav__link">
1. &nbsp; 贪心策略确定
</a>
</li>
<li class="md-nav__item">
<a href="#_2" class="md-nav__link">
代码实现
<a href="#2" class="md-nav__link">
2. &nbsp; 代码实现
</a>
</li>
<li class="md-nav__item">
<a href="#_3" class="md-nav__link">
正确性证明
<a href="#3" class="md-nav__link">
3. &nbsp; 正确性证明
</a>
</li>
@ -3449,7 +3449,7 @@
cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
\]</div>
<p>设数组长度为 <span class="arithmatex">\(n\)</span> ，两个隔板的组合数量（即状态总数）为 <span class="arithmatex">\(C_n^2 = \frac{n(n - 1)}{2}\)</span> 个。最直接地，<strong>我们可以穷举所有状态</strong>，从而求得最大容量，时间复杂度为 <span class="arithmatex">\(O(n^2)\)</span> 。</p>
<h3 id="_1">贪心策略确定<a class="headerlink" href="#_1" title="Permanent link">&para;</a></h3>
<h3 id="1">1. &nbsp; 贪心策略确定<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>这道题还有更高效率的解法。如下图所示，现选取一个状态 <span class="arithmatex">\([i, j]\)</span> ，其满足索引 <span class="arithmatex">\(i &lt; j\)</span> 且高度 <span class="arithmatex">\(ht[i] &lt; ht[j]\)</span> ，即 <span class="arithmatex">\(i\)</span> 为短板、 <span class="arithmatex">\(j\)</span> 为长板。</p>
<p><img alt="初始状态" src="../max_capacity_problem.assets/max_capacity_initial_state.png" /></p>
<p align="center"> 图：初始状态 </p>
@ -3506,7 +3506,7 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
</div>
<p align="center"> 图：最大容量问题的贪心过程 </p>
<h3 id="_2">代码实现<a class="headerlink" href="#_2" title="Permanent link">&para;</a></h3>
<h3 id="2">2. &nbsp; 代码实现<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>代码循环最多 <span class="arithmatex">\(n\)</span> 轮，<strong>因此时间复杂度为 <span class="arithmatex">\(O(n)\)</span></strong> 。</p>
<p>变量 <span class="arithmatex">\(i\)</span> , <span class="arithmatex">\(j\)</span> , <span class="arithmatex">\(res\)</span> 使用常数大小额外空间，<strong>因此空间复杂度为 <span class="arithmatex">\(O(1)\)</span></strong> 。</p>
<div class="tabbed-set tabbed-alternate" data-tabs="2:12"><input checked="checked" id="__tabbed_2_1" name="__tabbed_2" type="radio" /><input id="__tabbed_2_2" name="__tabbed_2" type="radio" /><input id="__tabbed_2_3" name="__tabbed_2" type="radio" /><input id="__tabbed_2_4" name="__tabbed_2" type="radio" /><input id="__tabbed_2_5" name="__tabbed_2" type="radio" /><input id="__tabbed_2_6" name="__tabbed_2" type="radio" /><input id="__tabbed_2_7" name="__tabbed_2" type="radio" /><input id="__tabbed_2_8" name="__tabbed_2" type="radio" /><input id="__tabbed_2_9" name="__tabbed_2" type="radio" /><input id="__tabbed_2_10" name="__tabbed_2" type="radio" /><input id="__tabbed_2_11" name="__tabbed_2" type="radio" /><input id="__tabbed_2_12" name="__tabbed_2" type="radio" /><div class="tabbed-labels"><label for="__tabbed_2_1">Java</label><label for="__tabbed_2_2">C++</label><label for="__tabbed_2_3">Python</label><label for="__tabbed_2_4">Go</label><label for="__tabbed_2_5">JS</label><label for="__tabbed_2_6">TS</label><label for="__tabbed_2_7">C</label><label for="__tabbed_2_8">C#</label><label for="__tabbed_2_9">Swift</label><label for="__tabbed_2_10">Zig</label><label for="__tabbed_2_11">Dart</label><label for="__tabbed_2_12">Rust</label></div>
@ -3692,7 +3692,7 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
</div>
</div>
</div>
<h3 id="_3">正确性证明<a class="headerlink" href="#_3" title="Permanent link">&para;</a></h3>
<h3 id="3">3. &nbsp; 正确性证明<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>之所以贪心比穷举更快，是因为每轮的贪心选择都会“跳过”一些状态。</p>
<p>比如在状态 <span class="arithmatex">\(cap[i, j]\)</span> 下，<span class="arithmatex">\(i\)</span> 为短板、<span class="arithmatex">\(j\)</span> 为长板。若贪心地将短板 <span class="arithmatex">\(i\)</span> 向内移动一格，会导致以下状态被“跳过”。<strong>这意味着之后无法验证这些状态的容量大小</strong>。</p>
<div class="arithmatex">\[