Levenshtein距離C＃計數錯誤類型

Question

我發現這段代碼，計算答案和猜測之間Levenshtein's distance：

int CheckErrors(string Answer, string Guess) 
{ 
    int[,] d = new int[Answer.Length + 1, Guess.Length + 1]; 
    for (int i = 0; i <= Answer.Length; i++) 
     d[i, 0] = i; 
    for (int j = 0; j <= Guess.Length; j++) 
     d[0, j] = j; 
    for (int j = 1; j <= Guess.Length; j++) 
     for (int i = 1; i <= Answer.Length; i++) 
      if (Answer[i - 1] == Guess[j - 1]) 
       d[i, j] = d[i - 1, j - 1]; //no operation 
      else 
       d[i, j] = Math.Min(Math.Min(
        d[i - 1, j] + 1, //a deletion 

        d[i, j - 1] + 1), //an insertion 

        d[i - 1, j - 1] + 1 //a substitution 

       ); 
    return d[Answer.Length, Guess.Length]; 
} 

但我需要一個方式做一個計數的時間發生的每個錯誤的數量。有沒有簡單的方法來實現呢？

Answer 1

好像你可以爲每個操作添加計數器：

   if (Answer[i - 1] == Guess[j - 1]) 
        d[i, j] = d[i - 1, j - 1]; //no operation 
       else 
       { 
        int del = d[i-1, j] + 1; 
        int ins = d[i, j-1] + 1; 
        int sub = d[i-1, j-1] + 1; 
        int op = Math.Min(Math.Min(del, ins), sub); 
        d[i, j] = op; 
        if (i == j) 
        { 
         if (op == del) 
          ++deletions; 
         else if (op == ins) 
          ++insertions; 
         else 
          ++substitutions; 
        } 
       }