LeetCode 72. Edit Distance--动态规划--Levenshtein Distance Algorithm--Java，Python解法

LeetCode 72. Edit Distance

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LeetCode题解专栏：LeetCode题解
LeetCode 所有题目总结：LeetCode 所有题目总结
大部分题目C++，Python，Java的解法都有。

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此题链接：Edit Distance - LeetCode
LeetCode 动态规划（Dynamic programming）系列题目:LeetCode 动态规划（Dynamic programming）系列题目
做这道题目前建议先做One Edit Distance - LeetCode
文章地址:LeetCode 161. One Edit Distance–Python,Java,C++解法 - zhangpeterx的博客 - CSDN博客

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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

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')

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这道题目是hard难度的，做法肯定是动态规划，至于怎么做就很难想了。
具体做法官方的结题文章写得很好：Edit Distance - LeetCode Articles

评论区里的大佬们发表了看法：

tomasnovella: Reaaaaally ? You guys don’t even mention in the solution that it’s a good old Levenshtein distance ?
Please if you use some known algorithm, at least mention it!

所以这道题目的英文正式名字叫做Levenshtein Distance Algorithm

Java官方解法如下：

class Solution {
  public int minDistance(String word1, String word2) {
    int n = word1.length();
    int m = word2.length();

    // if one of the strings is empty
    if (n * m == 0)
      return n + m;

    // array to store the convertion history
    int [][] d = new int[n + 1][m + 1];

    // init boundaries
    for (int i = 0; i < n + 1; i++) {
      d[i][0] = i;
    }
    for (int j = 0; j < m + 1; j++) {
      d[0][j] = j;
    }

    // DP compute 
    for (int i = 1; i < n + 1; i++) {
      for (int j = 1; j < m + 1; j++) {
        int left = d[i - 1][j] + 1;
        int down = d[i][j - 1] + 1;
        int left_down = d[i - 1][j - 1];
        if (word1.charAt(i - 1) != word2.charAt(j - 1))
          left_down += 1;
        d[i][j] = Math.min(left, Math.min(down, left_down));

      }
    }
    return d[n][m];
  }
}

Python官方解法如下：

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        n = len(word1)
        m = len(word2)
        
        # if one of the strings is empty
        if n * m == 0:
            return n + m
        
        # array to store the convertion history
        d = [ [0] * (m + 1) for _ in range(n + 1)]
        
        # init boundaries
        for i in range(n + 1):
            d[i][0] = i
        for j in range(m + 1):
            d[0][j] = j
        
        # DP compute 
        for i in range(1, n + 1):
            for j in range(1, m + 1):
                left = d[i - 1][j] + 1
                down = d[i][j - 1] + 1
                left_down = d[i - 1][j - 1] 
                if word1[i - 1] != word2[j - 1]:
                    left_down += 1
                d[i][j] = min(left, down, left_down)
        
        return d[n][m]