# 编辑距离
给你两个单词 word1 和 word2，请你计算出将 word1 转换成 word2 所使用的最少操作数 。
你可以对一个单词进行如下三种操作：
示例 1：
输入：word1 = "horse", word2 = "ros"
输出：3
解释：horse -> rorse (将 'h' 替换为 'r')rorse -> rose (删除 'r')rose -> ros (删除 'e')
示例 2：
输入：word1 = "intention", word2 = "execution"
输出：5
解释：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')
提示：
0 <= word1.length, word2.length <= 500word1 和 word2 由小写英文字母组成
以下程序实现了这一功能，请你填补空白处内容：
```python
class Solution(object):
def minDistance(self, word1, word2):
ls_1, ls_2 = len(word1), len(word2)
dp = list(range(ls_1 + 1))
for j in range(1, ls_2 + 1):
pre = dp[0]
dp[0] = j
for i in range(1, ls_1 + 1):
temp = dp[i]
if word1[i - 1] == word2[j - 1]:
dp[i] = pre
else:
____________________________
pre = temp
return dp[ls_1]
if __name__ == '__main__':
s = Solution()
print (s.minDistance("horse","ros"))
print (s.minDistance("intention","execution"))
```
## template
```python
class Solution(object):
def minDistance(self, word1, word2):
ls_1, ls_2 = len(word1), len(word2)
dp = list(range(ls_1 + 1))
for j in range(1, ls_2 + 1):
pre = dp[0]
dp[0] = j
for i in range(1, ls_1 + 1):
temp = dp[i]
if word1[i - 1] == word2[j - 1]:
dp[i] = pre
else:
dp[i] = min(pre + 1, dp[i] + 1, dp[i - 1] + 1)
pre = temp
return dp[ls_1]
if __name__ == '__main__':
s = Solution()
print (s.minDistance("horse","ros"))
print (s.minDistance("intention","execution"))
```
## 答案
```python
dp[i] = min(pre + 1, dp[i] + 1, dp[i - 1] + 1)
```
## 选项
### A
```python
dp[i] = min(pre + 1, dp[i] + 1, dp[i + 1] + 1)
```
### B
```python
dp[i] = min(pre + 1, dp[i] + 1, dp[i - 1] - 1)
```
### C
```python
dp[i] = min(pre + 1, dp[i] + 1, dp[i + 1] - 1)
```