LeetCode72 编辑距离（dp）

LeetCode72 编辑距离

对单词A删除一个等价于对单词B插入一个字符，基于此我们仅考虑如下三种操作：

• 对单词A插入一个字符
• 对单词B插入一个字符
• 对单词A替换一个字符

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        m, n = len(word1), len(word2)
        dp = [[505 for i in range(n + 1)] for j in range(m + 1)]

        for i in range(m + 1): dp[i][0] = i
        for i in range(n + 1): dp[0][i] = i

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if word1[i - 1] == word2[j - 1]:
                    dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i - 1][j - 1])
                else:
                    dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i - 1][j - 1] + 1)
        
        return dp[m][n]