Leetcode 72——编辑距离

题目链接

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

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        m, n = len(word1), len(word2)

        if m * n == 0:
            return m + n

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

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

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

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                similar = 0
                if word1[i - 1] != word2[j - 1]:
                    similar = 1
                dp[i][j] = min(dp[i - 1][j] + 1, min(dp[i][j - 1] + 1, dp[i - 1][j - 1] + similar))

        x, y = m, n
        steps = []
        while x > 0 or y > 0:
            tag = 1
            if word1[x - 1] == word2[y - 1]:
                tag = 0

            if dp[x - 1][y] == dp[x][y] - 1:
                steps.append((word1[x - 1], '_'))
                x = x - 1
            elif dp[x][y - 1] == dp[x][y] - 1:
                steps.append(('_', word2[y - 1]))
                y = x - 1
            elif dp[x - 1][y - 1] == dp[x][y] - tag:
                steps.append((word1[x - 1], word2[y - 1]))
                x, y = x - 1, y - 1
        steps.reverse()
        print(steps)
        return dp[m][n]