【leetcode】72. Edit Distance

题目如下：

解题思路：这题和【leetcode】583. Delete Operation for Two Strings也是相似的，只不过多了一种替换的操作。具体的解析可以参考《编程之美》中的《寻找相似字符串》章节，有详细的解释。不过书中给的解法是递归，在本题会TEL，所以需要改成动态规划。如果如果word1[i] != word2[j]，递推表达式变成：dp[i][j] = min(dp[i-1][j],dp[i][j-1],dp[i-1][j-1])+1。

代码如下：

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        if len(word1) == 0 or len(word2) == 0:
            return abs(len(word2) - len(word1))
        dp = [[0 for x in range(len(word2)+1)] for x in range(len(word1)+1)]
        for i in xrange(1,len(word1)+1):
            dp[i][0] = i
        for j in xrange(1,len(word2)+1):
            dp[0][j] = j
        for i in xrange(1,len(word1)+1):
            for j in xrange(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]