Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
You have the following 3 operations permitted on a word:
- Insert a character
- Delete a character
- Replace a character
Example 1:
Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')
Example 2:
Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')
Solution
这是一个典型的动态规划题目， 初始化第一行从0开始到len(word1)表示将一个空字符转化为word1需要的步骤。第一列为0开始到len(word2)表示将空字符转化为word2所需的步骤。 状态转移方程为：
dp[i][j]= dp[i-1][j-1] if word1[i]==word2[j] dp[i][j]= min(dp[i-1][j-1] , dp[i-1][j], dp[i][j-1]) + 1 if word1[i] !=word2[j]
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
m = len(word1)
n =len(word2)
dp=[[0 for j in range(n+1)] for i in range(m+1)]
for i in range(m+1):
dp[i][0] = i
for j in range(n+1):
dp[0][j] = j
for i in range(1,m+1):
for j in range(1,n+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], dp[i][j-1],dp[i-1][j-1])+1
return dp[m][n]
