leetcode 72. 编辑距离 Edit Distance (python)

目录

1. 代码
   1. 动规
   2. 递归(functools.lru_cache)

给定两个单词 word1 和 word2，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’)

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/edit-distance
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

代码

动规

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        n = len(word1)
        m = len(word2)
        dp = [[0] * (m+1) for _ in range(n+1)]
        for i in range(m+1): dp[0][i] = i
        for i in range(n+1): dp[i][0] = i
        for i in range(1, n+1):
            for j in range(1, m+1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1
        return dp[-1][-1]

递归(functools.lru_cache)

import functools

class Solution:
    @functools.lru_cache(None)
    def minDistance(self, word1: str, word2: str) -> int:
        if not word1 or not word2:
            return len(word1) + len(word2)
        if word1[-1] == word2[-1]:
            return self.minDistance(word1[:-1], word2[:-1])
        return min(self.minDistance(word1[:-1], word2), self.minDistance(word1, word2[:-1]), self.minDistance(word1[:-1], word2[:-1])) + 1