LeetCode 72 Edit Distance(Python详解及实现)

【题目】

Given two words word1 and word2, find theminimum number of steps required to convert word1 to word2. (each operation iscounted as 1 step.)

 

You have the following 3 operationspermitted on a word:

 

a) Insert a character

b) Delete a character

c) Replace a character

 

word1最少经过多少步可以变成word2。word1允许的操作如下：

a)插入一个字符；

b)删除一个字符；

c)替换一个字符。

比如“abc”变成“aba”只需要用"a"将最后一个字符"c"替换掉就可以。此时distance = 1

如word1[i] == word2[j],只需把word1[0至i-1]变为word2[0至j-1]。此时res[i][j] = res[i-1][j-1]

 

【思路】

l  替换：

若word1[i]替换为word2[j]是从word1到word2的最小改动，则需先把word1[0至i-1]以最小距离变为word2[0至j-1]，然后在执行替换操作，此时res[i][j] = res[i-1][j-1] + 1

l  插入：

若word1[i]插入某字符为是从word1到word2的最小改动，则需先把word1[0至i-1]以最小距离变为word2[0至j-1]，然后在执行插入操作，此时res[i][j] = res[i-1][j-1] + 1

l  删除：

若word1[i]删除某字符是从word1到word2的最小改动，则需先把word1[0至i-1]以最小距离变为word2[0至j-1]，然后在执行插入操作，此时res[i][j] = res[i-1][j-1] + 1

 

 

【Python实现】

class Solution:

    #@return an integer

   def minDistance(self, word1, word2):

       m=len(word1)+1

       n=len(word2)+1

       dp = [[0 for i in range(n)] for j in range(m)]#(m+1)*（n+1）二维矩阵

       for i in range(n):

           dp[0][i]=i

       for i in range(m):

           dp[i][0]=i

       for i in range(1,m):

           for j in range(1,n):

                dp[i][j]=min(dp[i-1][j]+1,dp[i][j-1]+1, dp[i-1][j-1]+(0 if word1[i-1]==word2[j-1] else 1))

       return dp[m-1][n-1]

   

s = Solution()

s.minDistance("bird","ibdr")