Edit Distance (or Levenshtein Distance) python solution for leetcode EPI 17.2

 1 class Solution:
 2     # @return an integer
 3     def minDistance(self, word1, word2):
 4         #http://www.cnblogs.com/zuoyuan/p/3773134.html
 5         m=len(word1)+1; n=len(word2)+1
 6         dp = [[0 for i in range(n)] for j in range(m)]
 7         delCost = insCost = subCost = 1        # The cost for each operation
 8         
 9         for i in range(m):
10             dp[i][0]=i
11         for j in range(n):
12             dp[0][j]=j
13         
14         for i in range(1,m):
15             for j in range(1,n):
16                 # del                      insert                      same                             sub
17                 #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))
18                 dp[i][j]=min(dp[i-1][j] + insCost, dp[i][j-1] + delCost, dp[i-1][j-1]+(0 if word1[i-1]==word2[j-1] else subCost))
19         return dp[m-1][n-1]
20         
21         """
22         解题思路：这道题是很有名的编辑距离问题。用动态规划来解决。
23         状态转移方程是这样的：dp[i][j]表示word1[0...i-1]到word2[0...j-1]的编辑距离。
24         而dp[i][0]显然等于i，因为只需要做i次删除操作就可以了。
25         同理dp[0][i]也是如此，等于i，因为只需做i次插入操作就可以了。
26         dp[i-1][j]变到dp[i][j]需要加1，因为word1[0...i-2]到word2[0...j-1]的距离是dp[i-1][j]，
27         而word1[0...i-1]到word1[0...i-2]需要执行一次删除，所以dp[i][j]=dp[i-1][j]+1；
28         同理dp[i][j]=dp[i][j-1]+1，因为还需要加一次word2的插入操作。
29         如果word[i-1]==word[j-1]，则dp[i][j]=dp[i-1][j-1]，
30         如果word[i-1]!=word[j-1]，那么需要执行一次替换replace操作，所以dp[i][j]=dp[i-1][j-1]+1，以上就是状态转移方程的推导。
31         """
32 ‘‘‘ Implement the minimum edit distance according to
33             the description of WikiPedia:
34             http://en.wikipedia.org/wiki/Edit_distance
35             Some other implements are available here:
36             http://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Levenshtein_distance
37         ‘‘‘
38         #if len(word1) < len(word2):
39         #    word1, word2 =word2, word1
40         """
41         delCost = insCost = subCost = 1        # The cost for each operation
42         steps = range(len(word1)+1)
43         
44         for word2Index in xrange(len(word2)):
45             diag = steps[0]                     # d(i-1, j-1)
46             steps[0] += 1
47             
48             for word1Index in xrange(len(word1)):
49                 temp = steps[word1Index+1]      # d(i-1, j)
50                 
51                 if word1[word1Index] == word2[word2Index]:
52                     steps[word1Index+1] = diag
53                 else:
54                     steps[word1Index+1] = min(steps[word1Index+1] + insCost, 55                                               steps[word1Index] + delCost, 56                                               diag + subCost)
57                 
58                 diag = temp                     # Store d(i-1, j-1) for next cell
59                     
60         return steps[-1]
61         """