[Backtracking/DP]72. Edit Distan

• 分类：Backtracking/DP
• 时间复杂度: O(m*n)

72. Edit Distance

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:

1. Insert a character
2. Delete a character
3. 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')

代码：

DP解法：

class Solution:
    def minDistance(self, word1: 'str', word2: 'str') -> 'int':
        
        m=len(word1)
        n=len(word2)
        
        dp=[[0 for i in range(n+1)] for i in range(m+1)]
        
        for i in range(1,m+1):
            dp[i][0]=dp[i-1][0]+1
        for j in range(1,n+1):
            dp[0][j]=dp[0][j-1]+1
            
        for i in range(1,m+1):
            for j in range(1,n+1):
                if (word1[i-1]==word2[j-1]):
                    dp[i][j]=dp[i-1][j-1]
                else:
                    #                 replace     insert      delete
                    dp[i][j]=min(min(dp[i-1][j-1],dp[i][j-1]),dp[i-1][j])+1
        
        return dp[-1][-1]

记忆化递归解法：

class Solution:
    def minDistance(self, word1: 'str', word2: 'str') -> 'int':
        
        m=len(word1)
        n=len(word2)
        
        res=self.paths(word1,word2,m,n,{})
        
        return res
    
    def paths(self, word1, word2, m, n, memo):   
        if (m,n) in memo:
            return memo[(m,n)]
        
        else:
            #退出条件
            if m==0 or n==0:
                memo[(m,n)]=max(m,n)
                return memo[(m,n)]

            if word1[m-1]==word2[n-1]:
                memo[(m,n)]=self.paths(word1,word2,m-1,n-1,memo)
            else:
                memo[(m,n)]=1+min(self.paths(word1,word2,m-1,n-1,memo),min(self.paths(word1,word2,m-1,n,memo),self.paths(word1,word2,m,n-1,memo)))
                
            return memo[(m,n)]

讨论：

1.记忆化递归比DP快
2.如果实在不能在一个循环里搞对的话弄两个循环也是OK的