72. Edit Distance

Description：

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

Insert a character
Delete a character
Replace a character

Example 1：

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2：

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

解题思路：

（1）递归解法

从两个字符串的最后的位置开始考虑:

如果最后两个字符(i,j)相等，最后两个字符就不要配对，所以等于minDistance(s1[0..i-1]，s2[0...j-1])；
如果最后两个字符不相等: 说明要编辑，具体可以分为三种情况:

a. 如果 s1[i-1]和s2[j]可以配对，那我就删除s1[i]即可(删除)；

b. 如果 s1[i]和s2[j-1]可以配对，那我就在s1的后面加上s2[j]即可(插入)；

c. 如果 s1[i-1]和s2[j-1]可以配对，那我就把s1[i]修改成s2[j]即可；

上面图片来源： https://blog.csdn.net/zxzxzx0119/article/details/82054807

class Solution:def minDistance(self, word1: str, word2: str) -> int:l1 = len(word1)l2 = len(word2)def tryMinDistance(i, j):if i == -1:return j + 1elif j == -1:return i + 1elif (word1[i] == word2[j]):return tryMinDistance(i - 1, j - 1)else:return (1 + min(tryMinDistance(i - 1, j - 1),  # replace itryMinDistance(i - 1, j),  # delete itryMinDistance(i, j - 1)  # add i+1 index))return tryMinDistance(l1 - 1, l2 - 1)so = Solution()
# test 1
word1 = "horse"
word2 = "ros"# test 2
# word1 = "intention"
# word2 = "execution"# test 3
# word1 = "dinitrophenylhydrazine"
# word2 = "benzalphenylhydrazone"# test 4
# word1 = ""
# word2 =""
print(so.minDistance(word1, word2))

（2）记忆化递归法

下面代码没有AC，代码有问题。

class Solution:def minDistance(self, word1: str, word2: str) -> int:l1 = len(word1)l2 = len(word2)dp = [[-1] * (l2)] * (l1)def tryMinDistance(i, j):if i == -1:return j + 1elif j == -1:return i + 1elif dp[i][j] != -1:return dp[i][j]elif (word1[i] == word2[j]):dp[i][j] = tryMinDistance(i-1, j-1)else:dp[i][j] = 1 + min(tryMinDistance(i-1, j-1),    # replace itryMinDistance(i-1, j),  # delete itryMinDistance(i, j-1)   # add i+1 index)print(dp)return dp[i][j]return tryMinDistance(l1-1, l2-1)so = Solution()
# test 1
word1 = "horse"
word2 = "ros"# test 2
# word1 = "intention"
# word2 = "execution"# test 3
# word1 = "dinitrophenylhydrazine"
# word2 = "benzalphenylhydrazone"# test 4
# word1 = ""
# word2 =""
print(so.minDistance(word1, word2))

上面代码总是在第一个测试用例中出错，得到结果是4。经过好长时间分析测试，发现是以下代码有问题：

dp = [[None] * (len(word2)) for _ in range(len(word1))]
print(dp)dp1 = [[None] * (len(word2))] * (len(word1))
print(dp1)

上面两种方式初始化列表，虽然得到结果形式上是一致的，却能造成不同的结果。目前还没想清楚，这两种方式的不同。

20190603补充：

上面两种初始化方式在更新列表数据时有差别，我们来看两个例子，感受一下：

例1：

dp = [[1] * 2] * 4
print(dp)
dp[0][1]=5
print(dp)

结果：

例2：

dp = [[1] * 2 for _ in range(4)]
print(dp)
dp[0][1]=5
print(dp)

结果：

已经AC的代码，如下：

class Solution:def minDistance(self, word1: str, word2: str) -> int:l1 = len(word1)l2 = len(word2)dp = dp = [[-1] * (len(word2)) for _ in range(len(word1))]def tryMinDistance(i, j):if i == -1:return j + 1elif j == -1:return i + 1elif dp[i][j] != -1:return dp[i][j]elif (word1[i] == word2[j]):dp[i][j] = tryMinDistance(i - 1, j - 1)else:dp[i][j] = 1 + min(tryMinDistance(i - 1, j - 1),  # replace itryMinDistance(i - 1, j),  # delete itryMinDistance(i, j - 1)  # add i+1 index)return dp[i][j]return tryMinDistance(l1 - 1, l2 - 1)so = Solution()
# test 1
# word1 = "horse"
# word2 = "ros"# test 2
# word1 = "intention"
# word2 = "execution"# test 3
# word1 = "dinitrophenylhydrazine"
# word2 = "benzalphenylhydrazone"# test 4
# word1 = ""
# word2 =""# test 5
word1 = "sea"
word2 = "eat"
print(so.minDistance(word1, word2))

（3）动态规划解法

上面图片来源：https://blog.csdn.net/zxzxzx0119/article/details/82054807

二维动态规划就是使用一个二维数组从左到右，从上到下来递归出最后的答案，注意几点:

dp数组的大小 dp[chs1.length + 1] [chs2.length + 1]，因为一开始是空串" "，所以要多开一个；
一开始初始化第一行和第一列，第一行表示的是chs1为空字符串""变成chs2要添加chs2长度的次数。第一列表示的是chs2为空字符串""变成chs1需要添加chs1长度的次数。

已经AC的代码：

class Solution:def minDistance(self, word1: str, word2: str) -> int:l1 = len(word1)l2 = len(word2)dp = [[0] * (l2+1) for _ in range(l1+1)]# print(dp)for i in range(l1+1):dp[i][0] = ifor j in range(l2+1):dp[0][j] = jfor i in range(1, l1+1):for j in range(1, l2+1):if word1[i - 1] == word2[j - 1]:c  = 0else:c = 1dp[i][j] = min(dp[i-1][j-1]+c, min(dp[i][j-1], dp[i-1][j])+1)return dp[l1][l2]

Reference：

【1】https://leetcode.com/problems/edit-distance/

【2】花花酱 LeetCode 72. Edit Distance - 刷题找工作 EP87

【3】LeetCode - 72. Edit Distance(编辑距离问题)(三个依赖的滚动优化)

【4】https://blog.csdn.net/zxzxzx0119/article/details/82054807

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【LeetCode】72. Edit Distance

72. Edit Distance

Description：

解题思路：

Reference：

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