72. Edit Distance

问题

给定两个单词，找出能把word1变为word2的最小操作个数，允许操作有：对单个字符插入/删除/替换。

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

思路

用dp[i][j]表示word1[:i]和word2[:j]之间的解（考虑到dp要放置0字符的状态）。dp方程为：
如果word1[i] == word2[j]，两字符相等，不需要任何操作，显然dp[i][j] = dp[i-1][j-1]。
否则，要在原来的基础上进行插入或替换，在dp[i-1][j]或dp[i][j-1]的基础插入，或者在dp[i-1][j-1]的基础替换，所以有dp[i][j] = min(dp[i - 1][j], dp[i][j - 1], dp[i-1][j-1]) + 1。

时间复杂度O(n*m)，空间复杂度O(n*m)

代码

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

        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[len(word1)][len(word2)]

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712. Minimum ASCII Delete Sum for Two Strings