72 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:
- 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')
解法:
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
len1,len2 = len(word1),len(word2)
if len1 == 0 or len2 == 0:
return max(len1,len2)
dp = [[0 for _ in range(len2+1)] for _ in range(len1+1)]
for i in range(1,len1+1):
dp[i][0] += dp[i-1][0]+1
for j in range(1,len2+1):
dp[0][j] += dp[0][j-1]+1
for i in range(1,len1+1):
for j in range(1,len2+1):
if word1[i-1] == word2[j-1]:
dp[i][j] = dp[i-1][j-1]
else:
dp[i][j] = min(dp[i-1][j]+1,dp[i-1][j-1]+1,dp[i][j-1]+1)
return dp[-1][-1]