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chapter_greedy/max_capacity_problem/index.html

@@ -3480,15 +3480,12 @@ 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">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。如图 15-8 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>这道题还有更高效率的解法。如图 15-8 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 15-8 &nbsp; 初始状态 </p>
 
-<p>如图 15-9 所示，<strong>若此时将长板 <span class="arithmatex">\(j\)</span> 向短板 <span class="arithmatex">\(i\)</span> 靠近，则容量一定变小</strong>。这是因为在移动长板 <span class="arithmatex">\(j\)</span> 后：</p>
-<ul>
-<li>宽度 <span class="arithmatex">\(j-i\)</span> 肯定变小。</li>
-<li>高度由短板决定，因此高度只可能不变（ <span class="arithmatex">\(i\)</span> 仍为短板）或变小（移动后的 <span class="arithmatex">\(j\)</span> 成为短板）。</li>
-</ul>
+<p>如图 15-9 所示，<strong>若此时将长板 <span class="arithmatex">\(j\)</span> 向短板 <span class="arithmatex">\(i\)</span> 靠近，则容量一定变小</strong>。</p>
+<p>这是因为在移动长板 <span class="arithmatex">\(j\)</span> 后，宽度 <span class="arithmatex">\(j-i\)</span> 肯定变小；而高度由短板决定，因此高度只可能不变（ <span class="arithmatex">\(i\)</span> 仍为短板）或变小（移动后的 <span class="arithmatex">\(j\)</span> 成为短板）。</p>
 <p><img alt="向内移动长板后的状态" src="../max_capacity_problem.assets/max_capacity_moving_long_board.png" /></p>
 <p align="center"> 图 15-9 &nbsp; 向内移动长板后的状态 </p>
 
@@ -3499,10 +3496,10 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
 <p>由此便可推出本题的贪心策略：初始化两指针分裂容器两端，每轮向内收缩短板对应的指针，直至两指针相遇。</p>
 <p>图 15-11 展示了贪心策略的执行过程。</p>
 <ol>
-<li>初始状态下，指针 <span class="arithmatex">\(i\)</span> , <span class="arithmatex">\(j\)</span> 分列与数组两端。</li>
+<li>初始状态下，指针 <span class="arithmatex">\(i\)</span> 和 <span class="arithmatex">\(j\)</span> 分列与数组两端。</li>
 <li>计算当前状态的容量 <span class="arithmatex">\(cap[i, j]\)</span> ，并更新最大容量。</li>
 <li>比较板 <span class="arithmatex">\(i\)</span> 和 板 <span class="arithmatex">\(j\)</span> 的高度，并将短板向内移动一格。</li>
-<li>循环执行第 <code>2.</code> , <code>3.</code> 步，直至 <span class="arithmatex">\(i\)</span> 和 <span class="arithmatex">\(j\)</span> 相遇时结束。</li>
+<li>循环执行第 <code>2.</code> 和 <code>3.</code> 步，直至 <span class="arithmatex">\(i\)</span> 和 <span class="arithmatex">\(j\)</span> 相遇时结束。</li>
 </ol>
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@@ -3539,7 +3536,7 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
 
 <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>
+<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>
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