给定两个单词 word1 和 word2,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:

插入一个字符
删除一个字符
替换一个字符
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')

 class Solution:
def minDistance(self, string1: str, string2: str) -> int:
m = len(string1)
n = len(string2)
f = [[None for _ in range(n+1)] for i in range(m+1)]
for i in range (m+1):
for j in range(n+1):
if i== 0:
f[i][j]=j
continue
if j == 0:
f[i][j] = i
continue
#insert delete replace
if string1[i-1] == string2[j-1]:
f[i][j]=f[i-1][j-1]
else:
f[i][j] = min(f[i-1][j],f[i][j-1],f[i-1][j-1]) + 1
return f[-1][-1]

参考 b站

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