Leetcode75 72编辑距离

72. 编辑距离

给你两个单词 word1 和 word2， 请返回将 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')

提示：

• 0 <= word1.length, word2.length <= 500
• word1 和 word2 由小写英文字母组成

Solution 1

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        # matrix[i][j] is the minDistance for word1[:i] and word2[:j]
        matrix = [[0 for _ in range(len(word2)+1)]
                  for _ in range(len(word1)+1)]
        for i in range(1, len(word1)+1):
            matrix[i][0] = i
        for j in range(1, len(word2)+1):
            matrix[0][j] = j
        for i in range(1, len(word1)+1):
            for j in range(1, len(word2)+1):
                a = matrix[i][j-1]+1
                b = matrix[i-1][j]+1
                if (word1[i-1] == word2[j-1]):
                    c = matrix[i-1][j-1]
                else:
                    c = matrix[i-1][j-1]+1
                matrix[i][j] = min(min(a, b), c)
        return matrix[len(word1)][len(word2)]