[LeetCode] 72. Edit Distance 编辑距离 @python

Description

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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

给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。

你可以对一个单词进行如下三种操作：

• 插入一个字符
• 删除一个字符
• 替换一个字符

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 1:

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')

Solution

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典型的动态规划问题。

首先定义状态矩阵，dp[m][n]，其中m为word1的长度+1，n为word2的长度+1，为什么+1？因为要考虑如果word1或word2为空的情况，后面可以看到。

定义dp[i][j]为word1中前 i 个字符组成的串，与word2中前j个字符组成的串的编辑距离。

插入操作：在word1的前 i 个字符后插入一个字符，使得插入的字符等于新加入的word2[j]。这里要考虑清楚，插入操作对于原word1字符来说，i是没有前进的，而对于word2来说是前进了一位然后两个字符串才相等的。所以此时是dp[i][j]=dp[i][j-1]+1。

删除操作：在word1的第 i−1​ 个字符后删除一个字符，使得删除后的字符串word[:i-1]与word2[:j]相同。这里要考虑清楚，删除操作对于原word2字符来说， j−1​ 是没有前进的，而对于word1来说是删除了一位然后两个字符串才相等的。所以此时是dp[i][j]=dp[i-1][j]+(0 or 1)。

ifi==0andj==0,dp[i][j]=0(1)

ifi==0andj>0,dp[i][j]=j(2)

ifi>0andj==0,dp[i][j]=i(3)

if0<i≤mand0<j≤n,dp[i][j]=min(dp[i][j−1]+1,dp[i−1][j]+1,dp[i−1][j−1]+(0or1))(4)

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#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Fri Apr 27 16:48:29 2018

@author: saul
"""

class Solution:

    def minDistance(self, word1, word2):
        m=len(word1)+1; n=len(word2)+1
        dp = [[0 for i in range(n)] for j in range(m)]

        for i in range(n):
            dp[0][i]=i
        for i in range(m):
            dp[i][0]=i
        for i in range(1,m):
            for j in range(1,n):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1

        return dp[m-1][n-1]

word1 = "intention"
word2 = "execution"
test = Solution()
print(test.minDistance(word1, word2))