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Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.
You have the following three operations permitted on a word:
- Insert a character
- Delete a character
- Replace a character
Example 1:
<pre>Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e') </pre>
Example 2:
<pre>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') </pre>
Constraints:
0 <= word1.length, word2.length <= 500word1andword2consist of lowercase English letters.
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题目大意：
求编辑两个字符串的最短距离。编辑操作含加删一个字符，替换一个字符。
解题思路：
求最值且涉及到字符串考虑用DP。递归式为
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2dp[i][j] = dp[i-1][j-1] if word1[i-1] == word[j-1]
= min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1, otherwise
解题步骤：
N/A
注意事项：
- 初始值先word2长度再word1.
- 初始化上和左边界，表示当一个字符串为空时，另一个字符串的编辑距离是其长度。
Python代码：
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15# dp[i][j] = dp[i-1][j-1] if word1[i-1] == word[j-1]
# = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1, otherwise
def minDistance(self, word1: str, word2: str) -> int:
dp = [[0 for _ in range(len(word2) + 1)] for _ in range(len(word1) + 1)]
for i in range(1, len(dp)):
dp[i][0] = i
for j in range(1, len(dp[0])):
dp[0][j] = j
for i in range(1, len(dp)):
for j in range(1, len(dp[0])):
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[-1][-1]
算法分析：
时间复杂度为O(nm)，空间复杂度O(nm)
