class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        # dp: 以i - 1结尾的word1和以j - 1结尾的word2相同需要的最小步数
        # 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]) + 1
        dp = [[0] * (len(word2) + 1) for _ in range(len(word1) + 1)]
        for j in range(len(word2) + 1):
            dp[0][j] = j
        for i in range(len(word1) + 1):
            dp[i][0] = i
        for i in range(1, len(word1) + 1):
            for j in range(1, len(word2) + 1):
                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]) + 1
        return dp[-1][-1]

画一个二维数组图 递推一下就很清楚了

dp数组表示以i - 1结尾的word1和j - 1结尾的word2相同需要的最小步数，

dp[0][0]表示word1和word2都是空的情况，所以初始化的时候dp[0][j]就要等于j，意思是把word2变成空需要的步数，同理dp[i][0]要等于i，表示把word1变成空需要的步数。

递推公式分成两部分，如果是遍历到两个字符相等了，那么就直接dp[i][j] = dp[i - 1][j - 1] 因为以i - 2结尾的word1和j - 2结尾的word2相同需要的步数和以i - 1的相等

如果不相等，那就说明要多一步，要么在dp[i][j - 1]上多一步或者dp[i - 1][j]上多一步，因为求最小的，所以取个min。

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        # dp: 以i - 1结尾的word1和以j - 1结尾的word2相同需要的最小步数
        # 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
        dp = [[0] * (len(word2) + 1) for _ in range(len(word1) + 1)]
        for j in range(len(word2) + 1):
            dp[0][j] = j
        for i in range(len(word1) + 1):
            dp[i][0] = i
        for i in range(1, len(word1) + 1):
            for j in range(1, len(word2) + 1):
                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]

删除或者替换的情况都考虑，就是多一个dp[i - 1][j - 1]的情况