# 编辑距离

<p>给你两个单词 <code>word1</code> 和 <code>word2</code>，请你计算出将 <code>word1</code> 转换成 <code>word2</code><em> </em>所使用的最少操作数 。</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> </p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong>word1 = "horse", word2 = "ros"<strong><br />输出：</strong>3<strong><br />解释：</strong>horse -> rorse (将 'h' 替换为 'r')rorse -> rose (删除 'r')rose -> ros (删除 'e')</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>word1 = "intention", word2 = "execution"<strong><br />输出：</strong>5<strong><br />解释：</strong>intention -> inention (删除 't')inention -> enention (将 'i' 替换为 'e')enention -> exention (将 'n' 替换为 'x')exention -> exection (将 'n' 替换为 'c')exection -> execution (插入 'u')</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>0 <= word1.length, word2.length <= 500</code></li>	<li><code>word1</code> 和 <code>word2</code> 由小写英文字母组成</li></ul>

以下程序实现了这一功能，请你填补空白处内容：

```java
class Solution {
	public int minDistance(String word1, String word2) {
		int len1 = word1.length();
		int len2 = word2.length();

		if (len1 * len2 == 0)
			return len1 + len2;
		String longerStr = len1 > len2 ? word1 : word2;
		String shorterStr = len1 > len2 ? word2 : word1;
		int shorterOne = Math.min(len1, len2);

		int[] dp = new int[shorterOne + 1];

		for (int i = 0; i < shorterOne + 1; i++) {
			dp[i] = i;
		}

		for (int j = 1; j <= longerStr.length(); j++) {

			int left = j;

			for (int i = 1; i <= shorterStr.length(); i++) {
				int updateDown = dp[i] + 1;
				int updateLeft = left + 1;
				int updateLeftDown = dp[i - 1];

				if (longerStr.charAt(j - 1) != shorterStr.charAt(i - 1)) {
					updateLeftDown++;
				}

				int min = Math.min(updateLeft, Math.min(updateDown, updateLeftDown));

				dp[i - 1] = left;

				_____________________;
			}
		}
		return dp[shorterOne];
	}
}

```

## template

```java
class Solution {
	public int minDistance(String word1, String word2) {
		int len1 = word1.length();
		int len2 = word2.length();

		if (len1 * len2 == 0)
			return len1 + len2;
		String longerStr = len1 > len2 ? word1 : word2;
		String shorterStr = len1 > len2 ? word2 : word1;
		int shorterOne = Math.min(len1, len2);

		int[] dp = new int[shorterOne + 1];

		for (int i = 0; i < shorterOne + 1; i++) {
			dp[i] = i;
		}

		for (int j = 1; j <= longerStr.length(); j++) {

			int left = j;

			for (int i = 1; i <= shorterStr.length(); i++) {
				int updateDown = dp[i] + 1;
				int updateLeft = left + 1;
				int updateLeftDown = dp[i - 1];

				if (longerStr.charAt(j - 1) != shorterStr.charAt(i - 1)) {
					updateLeftDown++;
				}

				int min = Math.min(updateLeft, Math.min(updateDown, updateLeftDown));

				dp[i - 1] = left;

				if (i == dp.length - 1) {
					dp[i] = min;
				} else {

					left = min;
				}
			}
		}
		return dp[shorterOne];
	}
}

```

## 答案

```java
if (i == dp.length - 1) {
	dp[i] = min;
} else {
	left = min;
}
```

## 选项

### A

```java
if (i == dp.length) {
	dp[i] = min;
} else {
	left = min;
}
```

### B

```java
if (i == dp.length + 1) {
	dp[i] = min;
} else {
	left = min;
}
```

### C

```java
if (i == dp.length - 1) {
	left = min;
} else {
	dp[i] = min;
}
```