feat(leecode): 72.编辑距离

docs/10.LeeCode/72.编辑距离.md

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+---
+title: 72.编辑距离
+date: 2024-03-30
+lang: 'zh-CN'
+sidebar: 'auto'
+categories:
+  - LeeCode
+tags:
+location: HangZhou
+---
+
+# Heading
+
+[[toc]]
+
+[72.编辑距离](https://leetcode.cn/problems/edit-distance/description/)
+
+Tags: algorithms string dynamic-programming
+
+Langs: c cpp csharp dart elixir erlang golang java javascript kotlin php python python3 racket ruby rust scala swift typescript
+
+- algorithms
+- Medium (62.84%)
+- Likes: 3347
+- Dislikes: -
+- Total Accepted: 463K
+- Total Submissions: 737.2K
+- Testcase Example: '"horse"\n"ros"'
+
+<p>给你两个单词&nbsp;<code>word1</code> 和&nbsp;<code>word2</code>， <em>请返回将&nbsp;<code>word1</code>&nbsp;转换成&nbsp;<code>word2</code> 所使用的最少操作数</em> &nbsp;。</p>
+
+<p>你可以对一个单词进行如下三种操作：</p>
+
+<ul>
+	<li>插入一个字符</li>
+	<li>删除一个字符</li>
+	<li>替换一个字符</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong>示例&nbsp;1：</strong></p>
+
+<pre>
+<strong>输入：</strong>word1 = "horse", word2 = "ros"
+<strong>输出：</strong>3
+<strong>解释：</strong>
+horse -&gt; rorse (将 'h' 替换为 'r')
+rorse -&gt; rose (删除 'r')
+rose -&gt; ros (删除 'e')
+</pre>
+
+<p><strong>示例&nbsp;2：</strong></p>
+
+<pre>
+<strong>输入：</strong>word1 = "intention", word2 = "execution"
+<strong>输出：</strong>5
+<strong>解释：</strong>
+intention -&gt; inention (删除 't')
+inention -&gt; enention (将 'i' 替换为 'e')
+enention -&gt; exention (将 'n' 替换为 'x')
+exention -&gt; exection (将 'n' 替换为 'c')
+exection -&gt; execution (插入 'u')
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>0 &lt;= word1.length, word2.length &lt;= 500</code></li>
+	<li><code>word1</code> 和 <code>word2</code> 由小写英文字母组成</li>
+</ul>
+
+<<< @/src/LeeCode/72.编辑距离.js

src/LeeCode/72.编辑距离.js

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+/*
+ * @lc app=leetcode.cn id=72 lang=javascript
+ *
+ * [72] 编辑距离
+ * 动态规划  Hard
+ */
+
+// @lc code=start
+/**
+ * @param {string} word1
+ * @param {string} word2
+ * @return {number}
+ */
+var minDistance = function (word1, word2) {
+  const dp = []
+  const m = word1.length
+  const n = word2.length
+
+  if (m * n === 0) {
+    return m + n
+  }
+
+  for (let i = 0; i < m + 1; i++) {
+    dp[i] = [i]
+  }
+  for (let j = 0; j < n + 1; j++) {
+    dp[0][j] = j
+  }
+
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      if (word1[i] === word2[j]) {
+        dp[i + 1][j + 1] = dp[i][j]
+      } else {
+        dp[i + 1][j + 1] = 1 + Math.min(dp[i][j], dp[i][j + 1], dp[i + 1][j])
+      }
+    }
+  }
+
+  return dp[m][n]
+}
+// @lc code=end