72. Edit Distance【H】【65】

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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

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问题：找出字符串的编辑距离，即把一个字符串s1最少经过多少步操作变成编程字符串s2，操作有三种，添加一个字符，删除一个字符，修改一个字符

解析：

首先定义这样一个函数——edit(i, j)，它表示第一个字符串的长度为i的子串到第二个字符串的长度为j的子串的编辑距离。

显然可以有如下动态规划公式：

if i == 0 且 j == 0，edit(i, j) = 0
if i == 0 且 j > 0，edit(i, j) = j
if i > 0 且j == 0，edit(i, j) = i

if i ≥ 1 且 j ≥ 1 ，edit(i, j) == min{ edit(i-1, j) + 1, edit(i, j-1) + 1, edit(i-1, j-1) + f(i, j) }，当第一个字符串的第i个字符不等于第二个字符串的第j个字符时，f(i, j) = 1；否则，f(i, j) = 0。

class Solution(object):    def minDistance(self, word1, word2):        s1 = word1        s2 = word2        l1 = len(s1) + 1        l2 = len(s2) + 1        if s1 == '' or s2 == '':            return max(l1,l2) - 1        t = [0]*l1        res =[0] * l2        for i in xrange(l2):            res[i] = t[:]        for i in xrange(l1):            res[0][i] = i        for i in xrange(l2):            res[i][0] = i        for i in xrange(1,l2):            for j in xrange(1,l1):                res[i][j] = min(res[i-1][j] + 1,res[i][j-1] + 1,res[i-1][j-1] + (s1[j - 1] != s2[i - 1 ]))        return res[-1][-1]

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