[Leetcode]Edit Distance

Given two 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

dp[i][j]表示word1前i个字母变成word2前j个字母的步数~如果word1的第i个字母等于word2的第j个字母, 则dp[i][j] = dp[i-1][j-1]。如果不等, 则有三种情况:

1) 把word1的前i-1个字母变成word2的前j-1个字母, 再把word1的第i个字母换成word2的第j个字母, 即dp[i-1][j-1] + 1

2) 把word1的前i个字母变成word2的前j-1个字母, 再加上word2的第j个字母, 即dp[i][j-1] + 1

3) 删掉word1的第i个字母, 把word1的前i-1个字母变成word2的前j个字母, 即1 + dp[i-1][j]

三种情况的最小值就是dp[i][j]

class Solution:    # @return an integer    def minDistance(self, word1, word2):        len1, len2 = len(word1), len(word2)        dp = [[0] * (len2 + 1) for i in xrange(len1 + 1)]        for i in xrange(len1 + 1):            dp[i][0] = i        for j in xrange(len2 + 1):            dp[0][j] = j        for i in xrange(1, len1 + 1):            for j in xrange(1, len2 + 1):                dp[i][j] = dp[i - 1][j - 1] if word1[i - 1] == word2[j - 1] else min(dp[i - 1][j - 1],                         dp[i - 1][j], dp[i][j - 1]) +  1        return dp[len1][len2]

还可以简化成一维的动态规划，可以参考http://blog.csdn.net/linhuanmars/article/details/24213795