就是两个都要删除,取两个都删的最低操作
python
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
dp=[[0]*(len(word2)+1) for _ in range(len(word1)+1)]
for i in range(len(word1)+1):
dp[i][0]=i
for j in range(len(word2)+1):
dp[0][j]=j
for i in range(1,len(word1)+1):
for j in range(1,len(word2)+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]+2,dp[i-1][j]+1,dp[i][j-1]+1)
return dp[-1][-1]
增删替--增==对面删 删==删 替:+1
python
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
dp=[[0]*(len(word2)+1) for _ in range(len(word1)+1)]
for i in range(len(word1)+1):
dp[i][0]=i
for j in range(len(word2)+1):
dp[0][j]=j
for i in range(1,len(word1)+1):
for j in range(1,len(word2)+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],dp[i][j-1])+1
return dp[-1][-1]
编辑距离正式结束