Description
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')
Solutions
1. DP-记忆化递归
还是比较难，得搞清楚 min 那里面对应的三个操作是什么。dp[i][j] 表示从 i 的左字串编辑到 j 个右字串最少需要的编辑次数。
# Time Complextiy: O(nm)
# Space Complextiy: O(nm)
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
l1, l2 = len(word1), len(word2)
self.dp = [[-1 for _ in range(l2+1)] for _ in range(l1+1)]
return self.min_dist(word1, word2, l1, l2)
def min_dist(self, word1, word2, l1, l2):
if l1 == 0:
return l2
if l2 == 0:
return l1
if self.dp[l1][l2] >= 0:
return self.dp[l1][l2]
if word1[l1-1] == word2[l2-1]:
ret = self.min_dist(word1, word2, l1-1, l2-1)
else:
ret = 1 + min(self.min_dist(word1, word2, l1-1, l2-1), # replace
self.min_dist(word1, word2, l1, l2-1), # add / insert
self.min_dist(word1, word2, l1-1, l2)) # delete
self.dp[l1][l2] = ret
return ret
# Runtime: 112 ms, faster than 93.16% of Python3 online submissions for Edit Distance.
# Memory Usage: 16.3 MB, less than 80.77% of Python3 online submissions for Edit Distance.
2. DP-迭代/递推
注意状态转移方程有两个部分！！！
# Time: O(mn)
# Space: O(mn)
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
n1, n2 = len(word1), len(word2)
dp = [[0 for _ in range(n2 + 1)] for _ in range(n1 + 1)]
dp[0][0] = 0
for i in range(1, n1 + 1):
dp[i][0] = dp[i - 1][0] + 1
for i in range(1, n2 + 1):
dp[0][i] = dp[0][i - 1] + 1
for i in range(1, n1 + 1):
for j in range(1, n2 + 1):
if word1[i - 1] == word2[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
else:
dp[i][j] = 1 + min(dp[i-1][j-1], # replace
dp[i-1][j], # delete
dp[i][j-1]) # insert
return dp[n1][n2]
# Runtime: 188 ms, faster than 60.15%
# Memory Usage: 16.4 MB, less than 80.77%