Edit Distance (or Levenshtein Distance) Python solution for Leetcode EPI 17.2

https://oj.leetcode.com/problems/edit-distance/

Edit Distance  

Given words word1 and word2, find the minimum number of steps required to convert word1 to Word2. (Each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

A) Insert a character
b) Delete a character
c) Replace a character

1 classSolution:2     #@return An integer3     defmindistance (self, Word1, Word2):4         #http://www.cnblogs.com/zuoyuan/p/3773134.html5M=len (word1) +1; N=len (WORD2) +16DP = [[0 forIinchRange (n)] forJinchrange (m)]7Delcost = Inscost = Subcost = 1#The cost of each operation8         9          forIinchRange (m):Tendp[i][0]=I One          forJinchrange (n): Adp[0][j]=J -          -          forIinchRange (1, m): the              forJinchRange (1, N): -                 #del Insert same sub -                 #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)) -Dp[i][j]=min (Dp[i-1][j] + inscost, dp[i][j-1] + delcost, dp[i-1][j-1]+ (0ifWORD1[I-1]==WORD2[J-1]Elsesubcost)) +         returnDp[m-1][n-1] -          +         """ A Problem Solving idea: This question is a very famous editing distance question. Use dynamic programming to solve.  at The state transition equation is this: Dp[i][j] represents the editing distance of word1[0...i-1] to word2[0...j-1].  - and dp[i][0] is obviously equal to I, because only need to do I delete operation is OK.  - The same is true for dp[0][i], equal to I, because I can only do the insertion operation.  - Dp[i-1][j] to dp[i][j] need to add 1, because word1[0...i-2] to word2[0...j-1] The distance is dp[i-1][j], - and Word1[0...i-1] to word1[0...i-2] need to perform a delete, so dp[i][j]=dp[i-1][j]+1; - The same dp[i][j]=dp[i][j-1]+1, because you also need to add a word2 insert operation.  in if word[i-1]==word[j-1], then dp[i][j]=dp[i-1][j-1], - if word[i-1]!=word[j-1], then you need to perform a replacement replace operation, so dp[i][j]=dp[i-1][j-1]+1, the above is the derivation of the state transition equation.  to         """ + " "Implement The minimum edit distance according to - The description of WikiPedia: the http://en.wikipedia.org/wiki/Edit_distance * Some Other implements is available here: $ http://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Levenshtein_distancePanax Notoginseng         " " -         #If Len (word1) < Len (word2): the         #word1, Word2 =word2, Word1 +         """ A delcost = Inscost = Subcost = 1 # The cost of each operation the steps = Range (len (word1) +1) +          - For Word2index in Xrange (Len (word2)): $ diag = steps[0] # d (i-1, j-1) $ Steps[0] + = 1 -              - For Word1index in Xrange (Len (word1)): the temp = steps[word1index+1] # d (I-1, J) -                 Wuyi if word1[word1index] = = Word2[word2index]: the steps[word1index+1] = diag - Else: Wu steps[word1index+1] = min (steps[word1index+1] + inscost, - Steps[word1index] + delcost, About diag + subcost) $                  - diag = temp # Store D (i-1, j-1) for next cell -                      - return steps[-1] A         """

Edit Distance (or Levenshtein Distance) Python solution for Leetcode EPI 17.2