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@@ -18,7 +18,7 @@
<link rel="prev" href="../unbounded_knapsack_problem/">
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<li class="md-nav__item">
<a href="../../chapter_backtracking/subset_sum_problem/" class="md-nav__link">
12.3. &nbsp; 子集和问题（New）
12.3. &nbsp; 子集和问题
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@@ -2026,6 +2028,20 @@
<li class="md-nav__item">
<a href="../summary/" class="md-nav__link">
13.7. &nbsp; 小结（New）
</a>
</li>
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@@ -2210,13 +2226,13 @@
<li>若 <span class="arithmatex">\(s[n-1]\)</span> 和 <span class="arithmatex">\(t[m-1]\)</span> 相同，我们可以直接跳过它们，接下来考虑 <span class="arithmatex">\(s[n-2]\)</span> 和 <span class="arithmatex">\(t[m-2]\)</span> ;</li>
<li>若 <span class="arithmatex">\(s[n-1]\)</span> 和 <span class="arithmatex">\(t[m-1]\)</span> 不同，我们需要对 <span class="arithmatex">\(s\)</span> 进行一次编辑（插入、删除、替换），使得两字符串尾部的字符相同，从而可以跳过它们，考虑规模更小的问题；</li>
</ul>
<p>也就是说，我们在字符串 <span class="arithmatex">\(s\)</span> 中进行的每一轮决策（编辑操作），都会使得 <span class="arithmatex">\(s\)</span> 和 <span class="arithmatex">\(t\)</span> 中剩余的待匹配字符发生变化。因此，状态定义为当前在 <span class="arithmatex">\(s\)</span> , <span class="arithmatex">\(t\)</span> 中考虑的第 <span class="arithmatex">\(i\)</span> , <span class="arithmatex">\(j\)</span> 个字符，记为 <span class="arithmatex">\([i, j]\)</span> 。</p>
<p>也就是说，我们在字符串 <span class="arithmatex">\(s\)</span> 中进行的每一轮决策（编辑操作），都会使得 <span class="arithmatex">\(s\)</span> 和 <span class="arithmatex">\(t\)</span> 中剩余的待匹配字符发生变化。因此，状态为当前在 <span class="arithmatex">\(s\)</span> , <span class="arithmatex">\(t\)</span> 中考虑的第 <span class="arithmatex">\(i\)</span> , <span class="arithmatex">\(j\)</span> 个字符，记为 <span class="arithmatex">\([i, j]\)</span> 。</p>
<p>状态 <span class="arithmatex">\([i, j]\)</span> 对应的子问题：<strong>将 <span class="arithmatex">\(s\)</span> 的前 <span class="arithmatex">\(i\)</span> 个字符更改为 <span class="arithmatex">\(t\)</span> 的前 <span class="arithmatex">\(j\)</span> 个字符所需的最少编辑步数</strong>。</p>
<p>至此得到一个尺寸为 <span class="arithmatex">\((i+1) \times (j+1)\)</span> 的二维 <span class="arithmatex">\(dp\)</span> 表。</p>
<p><strong>第二步：找出最优子结构，进而推导出状态转移方程</strong></p>
<p>考虑子问题 <span class="arithmatex">\(dp[i, j]\)</span> ，其对应的两个字符串的尾部字符为 <span class="arithmatex">\(s[i-1]\)</span> 和 <span class="arithmatex">\(t[j-1]\)</span> ，可根据不同编辑操作分为三种情况：</p>
<ol>
<li>在 <span class="arithmatex">\(s\)</span> 尾部添加 <span class="arithmatex">\(t[j-1]\)</span> ，则剩余子问题 <span class="arithmatex">\(dp[i, j-1]\)</span> ；</li>
<li>在 <span class="arithmatex">\(s[i-1]\)</span> 之后添加 <span class="arithmatex">\(t[j-1]\)</span> ，则剩余子问题 <span class="arithmatex">\(dp[i, j-1]\)</span> ；</li>
<li>删除 <span class="arithmatex">\(s[i-1]\)</span> ，则剩余子问题 <span class="arithmatex">\(dp[i-1, j]\)</span> ；</li>
<li>将 <span class="arithmatex">\(s[i-1]\)</span> 替换为 <span class="arithmatex">\(t[j-1]\)</span> ，则剩余子问题 <span class="arithmatex">\(dp[i-1, j-1]\)</span> ；</li>
</ol>
@@ -2617,13 +2633,13 @@ dp[i, j] = dp[i-1, j-1]
<a href="../../chapter_appendix/installation/" class="md-footer__link md-footer__link--next" aria-label="下一页: 14.1. &amp;nbsp; 编程环境安装" rel="next">
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