Edit Distance

最新推荐文章于 2022-04-26 09:23:54 发布

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最新推荐文章于 2022-04-26 09:23:54 发布

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分类专栏： c++/c

本文链接：https://blog.youkuaiyun.com/njust_ecjtu/article/details/36695107

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本文介绍了一种计算两个字符串之间编辑距离的动态规划算法。通过插入、删除或替换字符的操作，找到将一个字符串转换为另一个字符串所需的最小步骤数。文章详细解释了算法实现，并提供了具体的代码示例。

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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

此题编程之美有，其上使用的是递归。此处为dp!

 int minDistance(string word1, string word2)
    {
      
      if(word1.empty() && word2.empty())return 0;
      
      if(word1.empty())return word2.size();
      
      if(word2.empty())return word1.size();
    
      vector< vector<int> > dp(word1.size()+1,vector<int>(word2.size()+1));
      
       dp[0][0] = 0;
       
       for(int i = 1;i <= word1.size();i++)
       {
           dp[i][0] = i;
           
       }
       
        for(int i = 1;i <= word2.size();i++)
        {
            dp[0][i] = i;
        }
        
      for(int i = 1;i <= word1.size();i++)
      {
        for(int j = 1;j <= word2.size();j++)
        {
          if(word1[i-1] == word2[j-1])
          {
               dp[i][j] = dp[i-1][j-1] ; 
          }
          else
          {
              int temp_min = min(dp[i-1][j-1],dp[i][j-1]);//3种可能的情况 替换，后面俩种添加与删除都有可能！
              dp[i][j] = min(temp_min,dp[i-1][j]) + 1;
          }
        }
      }
        return dp[word1.size()][word2.size()];
    }