编辑距离

2024-05-16 02:05:04 363

有两个字符串。第一个字符串是源字符串，第二个字符串是目标字符串。在此程序中，我们必须查找将第一个字符串转换为第二个字符串所需的编辑次数。

字符串的编辑可以是：插入一些元素，从第一个字符串中删除某些内容，或修改某些内容以转换为第二个字符串。

输入输出

Input:
Two strings to compare.
string 1: Programming
string 2: Programs
Output:
Enter the initial string: Programming
Enter the final string: Programs
The number of changes required to convert Programming to Programs is 4

算法

editCount(initStr, finalStr, initLen, finalLen)

输入-初始和最终字符串及其长度。
输出-使initStr成为finalStr所需的编辑次数。

Begin
   if initLen = 0, then
      return finalLen
   if finalLen := 0, then
      return initLen

   if initStr[initLen - 1] = finalStr[finalLen - 1], then
      return editCount(initStr, finalStr, initLen – 1, finalLen - 1)
   answer := 1 + min of (editCount(initStr, finalStr, initLen , finalLen - 1)),
      (editCount(initStr, finalStr, initLen – 1, finalLen ),
      (editCount(initStr, finalStr, initLen – 1, finalLen - 1)
   return answer
End

示例

#include<iostream>
using namespace std;

int min(int x, int y, int z) {       //find smallest among three numbers
   if(x < y) {
      if(x < z)
         return x;
      else
         return z;
   }else {
      if(y < z)
         return y;
      else
         return z;
   }
}

int editCount(string initStr , string finalStr, int initLen, intfinalLen) {
   if (initLen == 0)       //when initial string is empty, add all characters of final string
      return finalLen;

   if (finalLen == 0)       //when final string is empty, delete all characters from initial string
      return initLen;

   //当最后一个字符匹配时，递归检查前一个字符
   if (initStr[initLen-1] == finalStr[finalLen-1])
      return editCount(initStr, finalStr, initLen-1, finalLen-1);

   //如果找不到最后一个字符，则递归检查插入，删除和更新操作
   return 1 + min ( editCount(initStr, finalStr, initLen, finalLen- 1),  // insert
 
      editCount(initStr, finalStr, initLen-1, finalLen), // delete
      editCount(initStr, finalStr, initLen-1, finalLen-1) // update
   );
}

int main() {
   string initStr;
   string finalStr;

   cout << "Enter the initial string: "; cin >> initStr;
   cout << "Enter the final string: "; cin >> finalStr;
   cout << "The number of changes required to convert " << initStr << " to " << finalStr;
   cout << " is " << editCount( initStr , finalStr, initStr.size(), finalStr.size()) << endl;
}

输出结果

Enter the initial string: Programming
Enter the final string: Programs
The number of changes required to convert Programming to Programs is 4

编辑距离

输入输出

算法

示例

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