leetcode 72 编辑距离

---

Contact me:

Blog -> https://cugtyt.github.io/blog/index
Email -> cugtyt@qq.com
GitHub -> Cugtyt@GitHub

---

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')

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