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
编辑距离，非常经典的二维DP题目，也是传说中的双序列问题。
首先定义最优解的表示，f[i][j]表示将word1前i 个字符转化为word2前j个字符需要的step数目(注意为了方便处理初始值，f[0][0] = 0表示空串的最小编辑距离，之后定义最优解之间的转换状态, 即：
f[i][j] = min(f[i][j-1]+1,f[i-1][j]+1,f[i-1][j-1]) (word1[i-1]==word2[j-1])
f[i][j] = min(f[i][j-1]+1,f[i-1][j]+1,f[i-1][j-1]+1) (word1[i-1]!=word2[j-1])
注意在f[i][j-1],f[i-1][j],f[i-1][j-1]都是可以经过1步或者0步到达f[i][j]的子状态(增加word2[j-1],删除word1[i-1],替换(或者不替换)得到f[i][j]。
这种直接的解法代码如下：
class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
if not word1 and not word2:
return 0
l1 = len(word1)
l2 = len(word2)
res = [[0 for i in xrange(l2+1)] for j in xrange(l1+1)]
for i in xrange(1,l1+1):
res[i][0] = i
for j in xrange(1,l2+1):
res[0][j] = j
for i in xrange(1,l1+1):
for j in xrange(1,l2+1):
res[i][j] = min(res[i-1][j],res[i][j-1])+1
if word1[i-1] == word2[j-1]:
res[i][j] = min(res[i][j],res[i-1][j-1])
else:
res[i][j] = min(res[i][j],res[i-1][j-1]+1)
return res[l1][l2]
可以发现这种解法时间复杂度是O(mn),空间复杂度也为O(mn)，但是实际计算转换状态时，只需要f[i-1][j],f[i][j-1],f[i-1][j-1]三个值，可以使用之前多次使用的滑动数组（滚动数组来解决）。因为在矩阵每一行从左到右处理，所以如果用一个数组res表示处理到j时，res[j-1]已经更新为f[i][j-1]，res[j]还没更新为f[i-1][j-1]，唯一需要解决的是res[i-1][j-1]。一个好的办法是使用一个单独的变量pre保存res[i-1][j-1]。在每次更新res[j]时先把 res[j]的历史值存储下来，作为下一次更新要使用的f[i-1][j-1]。另外为了减小空间复杂度可以使res为比较短的单词的长度加1，另外上述代码的三次min可以减少为1次min,即提前对f[i-1][j-1]的情况进行处理，代码如下：
class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
if not word1 and not word2:
return 0
l1 = len(word1)
l2 = len(word2)
if l1 <= l2: #word1 have the max len
word1,word2 = word2,word1
l1,l2 = l2,l1
res = range(l2+1)
for i in xrange(1,l1+1):
pre = res[0]
res[0] = i
for j in xrange(1,l2+1):
tmp = res[j] ＃缓存f[i-1][j-1]
if word1[i-1] != word2[j-1]: ＃提前处理
pre += 1
res[j] = min(res[j-1]+1,res[j]+1,pre)
pre = tmp
return res[l2]
这道题不少人直接使用f[i][j]=f[i-1][j-1](word1[i-1][j-1])和f[i][j]= min(f[i-1][j],f[i][j-1],f[i-1][j-1]+1)这种格式，但是前者的正确性证明我还没有看到特别好的解释。需要后续再想想。
简易版本滚动数组优化：
class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
if len(word1) == 0 or len(word2) == 0:
return max(len(word1), len(word2))
m = len(word1)
n = len(word2)
dp = [[0] * (n+1) for i in xrange(2)]
for i in xrange(1, n+1):
dp[0][i] = i
for i in xrange(1, m+1):
dp[i%2][0] = i
for j in xrange(1, n+1):
if word1[i-1] == word2[j-1]:
dp[i%2][j] = min([dp[(i-1)%2][j]+1, dp[i%2][j-1]+1, dp[(i-1)%2][j-1]])
else:
dp[i%2][j] = min([dp[(i-1)%2][j]+1, dp[i%2][j-1]+1, dp[(i-1)%2][j-1]+1])
return dp[m%2][n]
